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PHP - Populating Rsform Dropdown Box With Php

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Populating Rsform Dropdown Box With Php	View Content

Hello,
I'm trying to populate a dropdown box in a RSForm using this code:

//<code>
$db = JFactory::getDbo();
$db->setQuery("SELECT Bruel_ID FROM mpctz_rsform_bruels");
return $db->loadObjectList();
//</code>

However, it displays nothing in the box and some code outside of it (see attached file). Can anyone help?
Thanks,

Dani Attached Files APNAE.jpg 16.92KB 0 downloads

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Populating Dropdown

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Could anyone point me in the correct direction of how to go about excluding results from a drown down menu, Basically i'm calling everyone who's a contact to the user in the code below, id like to remove contacts who already have a bank account with the user, So basically its something along the lines of if the player_id is already in the table bank_accounts as PlayersID, then its excluded, Hope this makes sense and any help would be great, Thanks

Code: [Select]
<select name="contact1" class="maintablepstats" id="contact1">
<option value=""></option>
<?php
$sql = "SELECT player_id, contact_id, name, is_active FROM contacts as c JOIN players as p ON c.contact_id = p.id WHERE c.player_id = $playerID AND is_active = 1 ORDER BY name ASC";
$que = mysql_query($sql) or die(mysql_error());
while($list = mysql_fetch_array($que)) {
?>
<option value="<?php echo $list['contact_id'] ?>"><?php echo $list['name'] ?></option>
<?php } ?>
</select>

Populating The Dropdown From Db

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I'm trying to retrieve data from my DB and have it populate the dropdown values inside a form:

echo "<option value='0.00' " . ($array['roastturkey'] == '0.00' ? 'selected="selected"' : '') . ">0</option>";echo "<option value='1.00' " . ($array['roastturkey'] == '1.00' ? 'selected="selected"' : '') . ">1</option>";echo "<option value='2.00'> " . ($array['roastturkey'] == '2.00' ? 'selected="selected"' : '') . ">2</option>";echo "<option value='3.00'> " . ($array['roastturkey'] == '3.00' ? 'selected="selected"' : '') . ">3</option>";echo "<option value='4.00'> " . ($array['roastturkey'] == '4.00' ? 'selected="selected"' : '') . ">4</option>";echo "<option value='5.00'> " . ($array['roastturkey'] == '5.00' ? 'selected="selected"' : '') . ">5</option>";echo "<option value='6.00'> " . ($array['roastturkey'] == '6.00' ? 'selected="selected"' : '') . ">6</option>";echo "<option value='7.00'> " . ($array['roastturkey'] == '7.00' ? 'selected="selected"' : '') . ">7</option>";echo "<option value='8.00'> " . ($array['roastturkey'] == '8.00' ? 'selected="selected"' : '') . ">8</option>";echo "<option value='9.00'> " . ($array['roastturkey'] == '9.00' ? 'selected="selected"' : '') . ">9</option>";echo "<option value='10.00'> " . ($array['roastturkey'] == '10.00' ? 'selected="selected

I'm using this code, but cannot get the data that's inside the DB to populate the dropdown value.

I want it to get the info from the database so that the option that is saved in the database will be new default when the page is loaded.

If I don't change it to the previous info it will update the database with the default option rather then the actual option.

Populating Dropdown Boxes

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I am trying to figure out how to display member records after selecting it from the box. I've got the dropdown box working which retrieves the members name but cannot figure out how to populate that members details on the same page?

I'm using php/mysql and although I want to display the records I also want a user to update that record aswell, creating it for administrators to update accounts?

Any help would be appreciated

I've researched that ajax or javascript might be helpful but not totally familiar with them.

Dropdown Box Populating The Bgcolour Of A Table

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Hi,

Can someone help me with the following problem...

I am populating a html table with results from a mysql query. These results populate the 1st of four columns. The second column is a "RAG STATUS" dropdown menu - so GREEN/AMBER/RED. When this is selected I want it to change the colour of the corresponding row it is on.

I have had a play around but can only get it to change the colour of the first row, no matter which dropdown menu I change.

Code is below. If anything is not clear please let me know. Any help appreciated

Code: [Select]

<?php

    include ("commonTop.php");
include("dbvariables.php");
include("functions.php");
?>

<script type="text/javascript">
function changeBGCol(status)
{
document.getElementById("colour").bgColor=status;
}
</script>

<?php

$Checkout_ID = 2;

$result = mysql_query( "SELECT Task FROM Task T, Checkout C Where T.Checkout_ID = C.Checkout_ID and C.Checkout_ID=
$Checkout_ID" )
or die("SELECT Error: ".mysql_error());

$num_rows = mysql_num_rows($result);

echo "This will be sent to the following recepients $num_rows records.<P>";
echo "<table width=70% border=3>\n";
echo "<tr><th>Check</th><th>STATUS</th><th>JIRA</th><th>Comments</th></tr>";

while ($get_info = mysql_fetch_row($result)){

echo "<tr id=colour>\n";

foreach ($get_info as $field)

echo "\t<td><font face=arial size=1/>$field</font></td>\n";

echo "\t<td><select name='Status'>
                 <option value='None'>-- Choose --</option>
<option onclick='changeBGCol(this.value)' value='green' >GREEN</option>
<option onclick='changeBGCol(this.value)' value='orange'>AMBER</option>
<option onclick='changeBGCol(this.value)' value='red'> RED</option>

              </select></td>";
echo "\t<td><input type='text' name='JIRA'></td>";

echo "\t<td><input type='text' name='Comments'></td>";

echo "</tr>\n";

}
print "</table>\n";
?>
[code]
</code>

Dropdown Populating From Another Dropdown

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I am not a developer but I can modify code to work for me. The following code works on my test machine (Windows 10, IIS, PHP 7.4) but doesn't work on my website (cPanel, Some version of Linux, PHP 7.4). The two dropdowns are for State and City. You are supposed to be able to select the state and then select a city from that state then bring up a report for craft breweries in the city. When selecting State from the first dropdown, the page refreshes, the URL is correct with the reports.php?cat=<STATE> so $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other?

Selected code from reports.php

<?php
$cat=$_REQUEST['cat'];

?>
<SCRIPT language=JavaScript>
   <!--
   function reload(form)
   {
       var val=form.cat.options[form.cat.options.selectedIndex].value;
       self.location='reports.php?cat=' + val ;
   }

</script>

<?Php
   ///////// Getting the State from Mysql table for first list box//////////
   $Cyquer2="SELECT DISTINCT breweries.State, states.state FROM breweries INNER JOIN states ON breweries.State=states.abbr ORDER BY states.state";
   $Cyresult2 = mysqli_query($con, $Cyquer2);

   /////// for second drop down list we will check if State is selected else we will display all the cities/////
   if(isset($cat) and strlen($cat) > 0)
   {
       $Cyquer="SELECT DISTINCT City FROM breweries where State='" . $cat . "' order by City";
   }
   else
   {
       $Cyquer="SELECT DISTINCT City FROM breweries order by City";
   }

   $Cyresult1 = $con->query($Cyquer);

   echo "<form method=post action='brewerylistbycity.php'>";
       ////////// Starting of first drop downlist ///////// -->
       echo "<select name='cat' onchange=\"reload(this.form)\"><option value=''>Select State</option>";
           while($Cynoticia2 = mysqli_fetch_array($Cyresult2))
           {
               if($Cynoticia2['State']==@$cat)
               {
                   echo "<option selected value='$Cynoticia2[State]'>$Cynoticia2[state]</option>"."<BR>";

               }
               else
               {
                   echo "<option value='$Cynoticia2[State]'>$Cynoticia2[state]</option>";

               }
           }
       echo "</select>";
       echo "</br>";

       ////////// Starting of second drop downlist /////////
       echo "</br>";
       echo "<select name='subcat' selected=''><option value=''>Select City</option>";
           while($Cynoticia = $Cyresult1->fetch_assoc())
           {
               echo "<option value='$Cynoticia[City]'>$Cynoticia[City]</option>";
           }
       echo "</select></br>";

       //// End Form /////
       echo "<input type=submit value=Submit>";
   echo "</form>";
?>

Populating Dropdown Lists With Mysql , Then Submitting Form To New Table

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hello. i have an issue where the data stored with an image is not saving to a mysql table.
the image data is ok, just not the selections from the dropdown lists.

here is the code

<?php
include ('connect.php');

// Insert any new image into database
if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") {
$fileName = $_FILES['imagefile']['name'];
$fileSize = $_FILES['imagefile']['size'];
$fileType = $_FILES['imagefile']['type'];
$content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name']));
$jeweltype = $_POST['jeweltype'];
$jewelsize = $_POST['jewelsize_in'];
$jewelcolour = $_POST['jewelcolour_in'];
$jewelmaterial = $_POST['jewelmaterial_in'];
$jewelgender = $_POST['jewelgender_in'];

if(!get_magic_quotes_gpc()) {
$fileName = addslashes($fileName);
}

// Checking file size
if ($fileSize < 150000) {
mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) "
. "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')");
} else {
$err = "The Image file to too large!";
}
}

// Find out about latest image

$gotten = mysql_query("select * from jewel_images order by row_id desc");
$row = mysql_fetch_assoc($gotten);
$bytes = $row['content'];

// If this is the image request, send out the image

if ($_REQUEST['pic'] == 1) {
header("Content-type: $row[type];");
print $bytes;
}
?>
<html>
<head>
<title>Upload an image to a database</title>
</head>
<body>
<font color="#FF3333"><?php echo $err ?></font>
<table>
<form name="Upload" enctype="multipart/form-data" method="post">
<tr>
<td>Upload <input type="file" name="imagefile"><br />
Jewelery Type:
<select>
<?php
$sql="SELECT jeweltype FROM jeweltypes";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result)){
?>
<option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option>
<?php } ?>
</select>
<br />
Jewelery Size:
<select>
<?php
$sql="SELECT jewelsize FROM jewelsizes";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result)){
?>
<option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option>
<?php } ?>
</select>
<br />
Jewelery Colour:
<select>
<?php
$sql="SELECT jewelcolour FROM jewelcolours";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result)){
?>
<option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option>
<?php } ?>
</select>
<br />
Jewelery Material:
<select>
<?php
$sql="SELECT jewelmaterial FROM jewelmaterials";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result)){
?>
<option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option>
<?php } ?>
</select>
<br />
Jewelery Gender:
<select>
<?php
$sql="SELECT jewelgender FROM jewelgenders";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result)){
?>
<option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option>
<?php } ?>
</select>
<br />
<input type="submit" name="xsubmit" value="Upload">
</td>
</tr>
<tr>
<td>Latest Image</td>
</tr>
<tr>
<td><img src="?pic=1"></td>
</tr>
</form>
</table>
</body>
</html>

==============================================
here is the query
===============================================

<html>
<head><title>Your Page Title</title></head>
<body>
<?php
$database="josh_jewel";
mysql_connect ("localhost", "xxxxxxxxx", "yyyyyyyyyyyy");
@mysql_select_db($database) or die( "Unable to select database");
$result = mysql_query( "SELECT jewelcolour FROM jewel_images" )
or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
print "There are $num_rows records.<P>";
print "<table width=400 border=1>\n";
while ($get_info = mysql_fetch_row($result)){
print "<tr>\n";
foreach ($get_info as $field)
print "\t<td><font face=arial size=1/>$field</font></td>\n";
print "</tr>\n";
}
print "</table>\n";
?>
</body>
</html>

Please Help Me Populating The Table

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<?php
$con=mysql_connect("localhost","root","");
if(!$con)
{
die('could not connect' .mysql_error());
}
mysql_select_db("hrc_fault",$con);
$query = "SELECT * " .
"FROM fault_book";
$result = mysql_query($query, $con)
or die(mysql_error());
$num_movies = mysql_num_rows($result);
$registered_comp=<<<EOD
<h2><center>Registered Fault HRC</center></h2>
<table width="70%" border="1" cellpadding="2"
cellspacing="2" align="center">
<tr>
<th>Job Cd No</th>
<th>Date</th>
<th>Section</th>
<th>Item Description</th>
<th>Item Sl.No</th>
<th>Fault</th>
</tr>
EOD;
$fault_details = '';
while ($row = mysql_fetch_array($result))
{
$jc_no = $row['jc_no'];
$date = $row['date'];
$section = $row['section'];
$itm_des = $row['itm_des'];
$itm_slno = $row['itm_slno'];
$fault_brf = $row['fault_brf'];
$fault_details .=<<<RAM
<tr>
   <td>$jc_no</td>
   <td>$date</td>
   <td>$section</td>
   <td>$itm_des</td>
   <td>$itm_slno</td>
   <td>$fault_brf</td>
</tr>
RAM;
}
$movie_footer ="</table>";

$movie =<<<MOVIE
$jc_no
$date
$section
$itm_des
$itm_slno
$fault_brf
MOVIE;
echo "There are $num_movies complains in our database";
echo $movie;

?>

I m using above code but it only display the last record please debug the code

Populating List Box

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hi,

I have a form that has 2 lists, first one has the months from Jan to Dec which i am done with, for the second list i want to have years from current year for other say 5 years. Example, we are in 2020 so i should have the list populate 2020,2021,2022,2023,2024 and next year it should show 2021-2025, how to do that.

Here is what i have as HTML:

<html>
<head>
<link rel="stylesheet" type="text/css" href="reportstyle.css">
</head>

<body>
	
				<div class="form-wrapper">
					<form action="reportanalysis.php" method="post">
					
					<div class="form-item">
						<h4>Select a month:</h4>
						<select id="months" name="months" required="required">
						<option selected="selected" value="chose">Choose one option...</option>
						<option value="jan">Jan - 01</option>
						<option value="feb">Feb - 02</option>
						<option value="mar">Mar - 03</option>
						<option value="apr">Apr - 04</option>
						<option value="may">May - 05</option>
						<option value="jun">Jun - 06</option>
						<option value="jul">Jul - 07</option>
						<option value="aug">Aug - 08</option>
						<option value="sep">Sep - 09</option>
						<option value="oct">Oct - 10</option>
						<option value="nov">Nov - 11</option>
						<option value="dec">Dec - 12</option>
						</select>
					</div>
					
					<div class="form-item">
						<h4>Select a year:</h4>
						<select id="years" name="years" required="required">
						<option selected="selected" value="chose">Choose one option...</option>
						<option value="2020">2020</option>
						<option value="2021">2021</option>
						<option value="2022">2022</option>
						<option value="2023">2023</option>
						<option value="2024">2024</option>
						</select>
					</div>
					
					<div class="button-panel">
						<center><button type="submit" name="reporting" class="buttonstyle">Generate Graph Report</button></center>
					</div>
					
					<div class="reminder">
						<p><a href="home.php">Return to Main Menu</a>
					</div>
					
					</form>
				</div>
				

</body>
</html>

Edited April 3, 2020 by ramiwahdan
typo error

Dynamic Dropdown Question, Submit 2 Values From Dropdown To A Form??

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Pagination Populating Prev And Next

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Hi,
I'm trying to populate the previous and next links with an id from mysql. The code below works but also displays ids that do not exist in the database. I want the code to only show me the rows that exist not the ones that do not exist. For example, at this time I have 5 rows. The ids of the rows are 1, 2, 3, 4, 6 when I get to say, id 6 and click on next it displays id 7 instead of going back to 1 or greying out next - meaning there's no more to view. Can someone help?

<?php
include('connection.php');


if(isset($_GET['id'])){
$start = $_GET['id'];
}else{
$start = 0;
}
$sql = mysql_num_rows(mysql_query("SELECT * FROM thetable"));

    
$result = mysqli_query($con,$sql);

$rows = mysql_fetch_array($result);

echo $rows['thetable'];


if($start == 0){
echo "Previous &laquo;";
}else{
echo "<a href=\"./thepage.php?id=" . ($start - 1) . "\">&laquo; Previous </a>";
}

if($start == $sql-1){
echo "Next &raquo;";
}else{
echo "<a href=\"./thepage.php?id=" . ($start + 1) . "\">Next &raquo;</a>";
}
?>



<?php
// End while loop.


mysqli_close($con);

?>

Dynamically Populating An Array?

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I need to fill an array dynamically with all the data in the users table. Any help will be appreciated, thanks!

Populate Dropdown On The Basis Of Another Dropdown Value In Php Using Jquery Ajax

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I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time.

I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.

When i load ajax.php i get 2 error mesages:
Notice: Undefined index: machinery in C:\wamp64\www\gecko\ajaxfile.php on line *8*
Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp64\www\site\ajaxfile.php on line *16*

This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID.

Index.php

    <!doctype html>
      <?PHP
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  "Database" => 
    "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";
    ?>
    <html>
    <head>
    <meta charset="utf-8">
    <title>Untitled Document</title>
    </head>
    
    <body>
                <section id="formaT2" class="formaT2 formContent">
                <div class="row">
                    <div class="col-md-2 col-3 row-color remove-mob"></div>
                    <div class="col-md-5 col-9 bg-img" style="padding-left: 0; 
    padding-right: 0;">
                        <h1>Form</h1>
                        <div class="rest-text">
                           <div class="contactFrm">
                              <p class="statusMsg <?php echo 
    !empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
                               
                                <form action="connection.php" method="post">
                                    <div>machinery</div>     
    <select id="machinery">
       <option value="0">--Please Select Machinery--</option>
       <?php 
       // Fetch Department
       $sql = "SELECT Machinery FROM T013";
       $machanery_data = sqlsrv_query($conn2,$sql);
       while($row = sqlsrv_fetch_array($machanery_data) ){
          $id = $row['Id'];
          $machinery = $row['Machinery'];
          
          // Option
          echo "<option value='".$id."' >".$machinery."</option>";
       }
       ?>
    </select>
    <div class="clear"></div>
                                    <div>Sub Machinery</div>
    <select id="sub_machinery">
       <option value="0">- Select -</option>
    </select>
                                    
                                    <input type="submit" name="submit" 
    id="submit" class="strelka-send" value="Insert">
                                    <div class="clear"> </div>
                                </form>
                             </div> 
                         </div>
                    </div>
                </div> 
            </section>
             </script>
          <script type="text/javascript">
          $(document).ready(function(){

        $("#machinery").change(function(){
            var machinery_id = $(this).val();
    
            $.ajax({
                url:'ajaxfile.php',
                type: 'post',
                data: {machinery:machinery_id},
                dataType: 'json',
                success:function(response){
    
                    var len = response.length;
     
                    $("#sub_machinery").empty();
                    for( var i = 0; i<len; i++){
                        var machinery_id = response[i]['machinery_id'];
                        var machinery = response[i]['machinery'];
                        
                        $("#sub_machinery").append("<option 
    value='"+machinery_id+"'>"+machinery+"</option>");
    
                    }
                }
            });
        });
    
    });
          </script>
    </body>
    </html>

Ajaxfile.php

<?php
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  
    "Database" => "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";

    $machineryID = $_POST['machinery'];   // department id

    $sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
    
    $result = sqlsrv_query($conn2,$sql);
    
    $machinery_arr = array();
    
    while( $row = sqlsrv_fetch_array($result) ){
        $machinery_id = $row['ID'];
        $machinery = $row['MachineID'];
    
        $machinery_arr[] = array("ID" => $machinery_id, "MachineID" => 
    $machinery);
    }
    
    // encoding array to json format
    echo json_encode($machinery_arr);
    ?>

Edited May 6, 2019 by davidd

Stuck, On Populating Multiple Lists

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Ok, so i got this website project dropped in my lap, and I'm stuck not really having done much php coding ever. I have a database which is linked already, and it was working when pulling data from a specific table. The problem arose when I was asked to make it work with other tables. I have a table with the table names in it, and the drop down field to select the table populates. The next field is a multiple select field that is supposed to populate report dates from the previously selected table, and then finally the last multiple select field is supposed to populate with names that match the dates selected from the previous field. The last one I can live without, as the existing code takes into account if the name selected doesn't exist in the table with the selected date, but at the moment the date field just comes up blank. Any help would be greatly appreciated.

Populating A Table With Gaps In The Sql Results

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How would I force a certain table format and then populate it with the SQL results in their respective columns and rows?

i.e.

Code: [Select]
+------------------+----------+----------+----------+
| | Col 1 | Col 2 | Col 3 |
+-------------------+----------+----------+----------+
| row 1 lab data | 16777216 | | 573 |
+-------------------+----------+----------+----------+
| row 2 lab data | 23454235 | 87247247 | 65743 |
+-------------------+----------+----------+----------+
| row 3 lab data | 16777216 | 47832364 | |
+-------------------+----------+----------+----------+
Currently I'm just using a while loop to populate it horizontally, but that doesn't work for row 1.

Code: [Select]
<?PHP
while($row = mysql_fetch_array($result)
{
?> blah blah blah html here <?PHP echo $row[0]; ?> more html <?PHP echo $row[1]; ?>
and so forth
<?PHP } ?>

It ends up producing this:

Code: [Select]

+------------------+----------+----------+----------+
| | Col 1 | Col 2 | Col 3 |
+-------------------+----------+----------+----------+
| row 1 lab data | 16777216 | 573 | |
+-------------------+----------+----------+----------+
| row 2 lab data | 23454235 | 87247247 | 65743 |
+-------------------+----------+----------+----------+
| row 3 lab data | 16777216 | 47832364 | |
+-------------------+----------+----------+----------+

Populating Of Drop Down Menu Not Working

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trying to get a drop-down menu (in a form) to work that needs to get populated from a table
table called : CarCategory
in this table are 2 columns called
CarCategoryID (unique number from 1-18)
CarCategoryDesc (actual name of category.. e.g. 'convertible')

link : http://98.131.37.90/postPart.php

I'm a total novice.. and can't figure out why the first is not working and the second (country) is working

code :


<label>
<div align="center">
<select name="carcategory" id="CarCategory" class="validate[required]" style="width: 200px;">

<option value="">Select Car Category from list...</option>

<?php while($obj_queryCarCategory = mysql_fetch_object($result_queryCarCategory)) { ?>
<option value="<?php echo $obj_queryCarCategory->CarCategoryID;?>"
<?php if($obj_queryCarCategory->CarCategoryID == $UserCountry) { echo 'selected="selected"'; } ?> >
<?php echo $obj_queryCarCategory->CarCategoryDesc;?></option>
<?php } ?>

</select>
</div>
</label>


<label>
<div align="center">
<select name="country" id="Country" class="validate[required]" style="width: 200px;">

<option value="">Select...</option>

<?php while($obj_queryCountry = mysql_fetch_object($result_queryCountry)) { ?>
<option value="<?php echo $obj_queryCountry->CountryID;?>"
<?php if($obj_queryCountry->CountryID == $UserCountry) { echo 'selected="selected"'; } ?> >
<?php echo $obj_queryCountry->CountryName;?></option>
<?php } ?>

</select>
</div>
</label>

Question About How To Populating A Sub Drop Down From An Other One With Php Code

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Hi!! the question look simple at his base because I will receive TONS of answer saying ... " hey men use javascript OR ajax " but I don't want to use these... I want to populate the second dropdown menu dynamicly with PHP code. and sql query.

why?? because I have over 43000 possibility and doing a script manually is impossible. the database is changing every hour.

this is the code of my working page.

the form

NOTE: i REMOVE ALL DB CONNECTION FOR SECURITY PURPOSE. THANKS FOR YOUR UNDERTANDING.

<!doctype html public "-//w3c//dtd html 3.2//en">

<html>

<head>
<title>(Type a title for your page here)</title>
</head>

<body >
<img src="http://www.scale24-25.com/images/banners/banner.jpg" alt="banner" width="997" height="213" border="0" usemap="#Map" />
<map name="Map" id="Map">
<area shape="poly" coords="0,100,85,102,92,55,108,50,326,51,323,7,3,3" href="http://www.hobby-shop.qc.ca" alt="hobby-shop" />
<area shape="poly" coords="672,57,891,53,887,4,995,4,985,108,673,102" href="http://www.hobby-shop.qc.ca" alt="hobby-shop" />
<area shape="poly" coords="326,13,330,53,98,54,90,92,670,99,668,54,889,51,886,5" href="http://www.scale-auto-style.com" alt="" />
</map>
<?php

if (isset($_POST['todo']) && $_POST['todo'] == "search") {    $todo=$_POST['todo'];

$manufacturer_reel=$_POST['manufacturer_reel'];

$query="select * from kit where  ";

////////// Including manufacturer_reel field search ////
if(strlen($manufacturer_reel) > 0 ){
$query.= " manufacturer_reel='$manufacturer_reel' and ";
}
//// End of class field search ///////////

$query=substr($query,0,(strLen($query)-4));

echo $query;
echo "<br><br>";
$nt=mysql_query($query);
echo mysql_error();

// End if form submitted

}else{

echo "<form method=post action='search-keyword.php?go'><input type=hidden name=todo value=search>";
?>
<table width="960" border="0">
  <tr>
    <td colspan="3"><p align="center"></td>
  </tr>
  <tr>
    <td><div align="right"></div></td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td><div align="right">Echelle du modele</div></td>
    <td align="center">
<?php
//  strat of drop down  ///
$q=mysql_query("SELECT DISTINCT scale FROM kit ORDER BY scale");
echo '<select name="scale"><option value="scale">Please Choose an Option</option>';
while($row = mysql_fetch_array($q)) {
$thing = $row['scale'];
echo '<option>'.$thing.'</option>';
}
echo "</select>";

echo " <br> \n";
echo " <br> \n";

?></td>
    <td><div align="left">Scale of model</div></td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
    <td><div align="right"></div></td>
  </tr>
  <tr>
    <td><div align="right">Nom du manufacturier automobile</div></td>
    <td align="center"><?php
//  strat of drop down  ///
$q=mysql_query("SELECT DISTINCT manufacturer_reel FROM kit ORDER BY manufacturer_reel");
echo '<select name="manufacturer_reel"><option value="reel manufacturer">Please Choose an Option</option>';
while($row = mysql_fetch_array($q)) {
$thing = $row['manufacturer_reel'];
echo '<option>'.$thing.'</option>';
}
echo "</select>";
//  end of drop down  ///
echo " <br> \n";
echo " <br> \n";
//  end of drop down  ///

?></td>
    <td><div align="left">Name Of vehicule manufacturer</div></td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
    <td><div align="right"></div></td>
  </tr>
  <tr>
    <td><div align="right">Nom du manufacturier de modele reduit</div></td>
    <td align="center"><?php
//  start of drop down  ///
$q=mysql_query("SELECT DISTINCT manufacturer_kit FROM kit ORDER BY manufacturer_kit");
echo '<select name="manufacturer_kit"><option value="kit manufacturer">Please Choose an Option</option>';
while($row = mysql_fetch_array($q)) {
$thing = $row['manufacturer_kit'];
echo '<option>'.$thing.'</option>';
}
echo "</select>";

//  end of drop down  ///
echo " <br> \n";
echo " <br> \n";

?></td>
    <td><div align="left">Name Of kit manufacturer</div></td>
  </tr>
  <tr>
    <td><div align="right"></div></td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td><div align="right">Annee de fabrication du Vehicule</div></td>
    <td align="center"><?php
//  strat of drop down  ///
$q=mysql_query("SELECT DISTINCT year_prod_reel FROM kit ORDER BY year_prod_reel");
echo '<select name="year_prod_reel"><option value="vehicules year manufactured">Please Choose an Option</option>';
while($row = mysql_fetch_array($q)) {
$thing = $row['year_prod_reel'];
echo '<option>'.$thing.'</option>';
}
echo "</select>";
//  end of drop down  ///
echo " <br> \n";
echo " <br> \n";

?></td>
    <td><div align="left">Year of production of the vehicule</div></td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td> </td>
    <td align="center"> </td>
    <td> </td>
  </tr>
  <tr>
    <td><div align="right"><strong></strong></div></td>
    <td align="center">
<?php

echo "
<br><input type=submit value=Search name='name' action='search-keyword.php?go' >
</form>
";
}

?>
<td><div align="left"><strong></strong></div></td>

</body>

</html>

AND THE PAGE RESULT

<!doctype html public "-//w3c//dtd html 3.2//en">

<html>

<head>
<title>(Type a title for your page here)</title>

</head>

<body >
<p align="center"><img src="http://www.scale24-25.com/catalog/images/banners/banner.jpg" alt="banner" width="997" height="213" border="0" usemap="#Map" />
<map name="Map" id="Map">
<area shape="poly" coords="0,100,85,102,92,55,108,50,326,51,323,7,3,3" href="http://www.hobby-shop.qc.ca" alt="hobby-shop" />
<area shape="poly" coords="672,57,891,53,887,4,995,4,985,108,673,102" href="http://www.hobby-shop.qc.ca" alt="hobby-shop" />
<area shape="poly" coords="325,11,329,51,97,52,89,90,669,97,667,52,888,49,885,3" href="http://www.scale-auto-style.com" alt="" />
</map></p>
<table width="920" border="0">
  <tr>
    <td width="359"><p class="style5">You will find some TBC. TBC is for "To Be Confirmed". If somebody know the missing information ( TBC ) it will be appreciate to send it to me at this mail: <a href="mailto:info@hobby-shop.qc.ca">info@hobby-shop.qc.ca</a> <br />
        <br />
thanks in advance for your help</p>
    </td>
    <td width="161"> </td>
    <td width="386"> <div align="right">
      <p>This logo <img  src= 'http://www.scale24-25.com/images/PDF_logo.gif' width='50' height='50'/> mean that you can download a PDF file. To read it you need minimum Acrobat reader and you can download it <a href="http://get.adobe.com/reader/">here</a></p>
      <p>Some link will return a TBC. Sorry for any inconveniant. as soon as we will receive or do the PDF it will be updated.</p>
    </div></td>
  </tr>
</table>

<br>

<?php
if (isset($_POST['todo']) && $_POST['todo'] == "search") {    $todo=$_POST['todo'];

$name=$_POST['manufacturer_reel'];
$name1=$_POST['manufacturer_kit'];
$name2=$_POST['year_prod_reel'];
$name3=$_POST['scale'];

//-query  the database table for the post field

$sql="SELECT  kit_id, kit_number, kit_name, description, manufacturer_kit, manufacturer_reel, scale, engine_detail, year_prod_kit, year_prod_reel, image_id, image_id1, image_id2  FROM kit ";
    $sqlOperand="WHERE";
    if ($name != "reel manufacturer") {
        $sql = $sql . $sqlOperand . " manufacturer_reel LIKE '%" . $name ."%' ";
        $sqlOperand = " AND ";
    }
    if ($name1 != "kit manufacturer") {
        $sql = $sql . $sqlOperand . " manufacturer_kit LIKE '%" . $name1 ."%' ";
        $sqlOperand = " AND ";
    }
    if ($name2 != "vehicules year manufactured") {
        $sql = $sql . $sqlOperand . " year_prod_reel LIKE '%" . $name2 ."%' ";
        $sqlOperand = " AND ";
    }
    if ($name3 != "scale") {
        $sql = $sql . $sqlOperand . " scale LIKE '%" . $name3 ."%' ";
        $sqlOperand = " AND ";
    }
    if ($sqlOperand == "WHERE") {
           echo  "<p>Please enter a search query</p>";
    }

else {
   //   echo '<img src="web-design/sorry.jpg" alt="sorry">';

       // echo "Sorry nothing match your selection / Desolle rien ne correspond a votre selection";
   // echo "run sql: <BR> $sql <BR> and display results";
    }

  //-run  the query against the mysql query function
   $result=mysql_query($sql);

  //-make the header of the table

   echo "  <table align=center width=\"1340\" border=\"\">\n";
   echo "    <tr>\n";
   echo"  <td nowrap align=center width='95'>  kit manufacturer  \n";
   echo"  <td nowrap align=center width='55'> scale \n";
   echo"  <td nowrap align=center width='90'> kit number \n";
   echo"  <td nowrap align=center width='290'> kit name \n";
   echo"  <td nowrap align=center width='400'> description \n";
   echo"  <td nowrap align=center width='125'> vehicule year \n";
   echo"  <td nowrap align=center width='80'> complete engine detail\n";
   echo"  <td nowrap align=center width='80'> Select Picture to get bigger\n";
   echo"  <td nowrap align=center width='75'> instruction sheet\n";
   echo"  <td nowrap align=center width='75'> box contain\n";
   echo "    <tr>\n";

   //-create  while loop and loop through result set
   while($row=mysql_fetch_array($result)){
         $description  =$row['description'];
         $manufacturer_kit=$row['manufacturer_kit'];
$scale=$row['scale'];
$manufacturer_reel=$row['manufacturer_reel'];
         $kit_id=$row['kit_id'];
$kit_number=$row['kit_number'];
$kit_name=$row['kit_name'];
$engine_detail=$row['engine_detail'];
$year_prod_reel=$row['year_prod_reel'];
$image_id=$row['image_id'];
$image_id1=$row['image_id1'];
$image_id2=$row['image_id2'];


   //-create  table of item during he while loop

echo "  <table align=center width=\"1340\" border=\"\">\n";
echo"  <td nowrap align=center width='95'>  $manufacturer_kit  \n";
echo"  <td nowrap align=center width='55'>  $scale  \n";
echo"  <td nowrap align=center width='90'> $kit_number \n";
echo"  <td nowrap align=center width='290'> $kit_name \n";
echo"  <td nowrap width='400'> $description \n";
echo"  <td nowrap align=center width='125'> $year_prod_reel \n";
echo"  <td nowrap align=center width='80'> $engine_detail \n";
echo"  <td nowrap width='80'> <a href=".$image_id.">   <img src='".$image_id."' width='75' height='50' border='0'/>";
echo"  <td nowrap width='75'> <a href= ".$image_id1.">   <img src= 'http://www.scale24-25.com/images/PDF_logo.gif' width='50' height='50' border='0'/>";
echo"  <td nowrap width='75'> <a href=".$image_id2.">   <img src='http://www.scale24-25.com/images/PDF_logo.gif' width='50' height='50' border='0'/>";
echo "    </tr>\n";
echo "    </table>\n";
echo "    </tr>\n";
echo "    </table>\n";

   }
   }
   else{
   echo  "<p>Please enter a search query</p>";
   }

?>



</body>

</html>

YOU CAN SEE IN THE FORM i HAVE A DROP DOWN MENU CALL Name Of vehicule manufacturer .. IN VACT THIS IS THE REAL CAR MANUFACTURER EX: TOYOTA, gmc ETC.. WHEN THE USER WILL SELECT ONE OF THESE AN OTHER DROPDOWN ( NOT SHOWN ) WILL DISPLAY EVERY MODEL FOUND IN THE DATABASE IN RELATION WITH THE MANUFACTURE. IN FACT IT'S LOCATED IN THE TABLE.

SO HOW I CAN GET THIS RELATION??

YOURS

SEBASTIEN

Populating A Drop Down Box From Values In Databases - Help

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Basically,

I have a database table called 'users' and I would like to populate a drop down box with these values of 'users'.

How?? - to call upon the values is this: 'upduser2'

Right now, all I am using is a text box, in where you have to type in the users name manually (this is so an admin can change variables and settings according to that current user). This is what I am using so far:

<h3>Update User Level</h3>
                                                    <? echo $form->error("upduser"); ?>
                                                    <table>
                                                      <form action="adminprocess.php" method="post">
                                                        <tr>
                                                          <td> Username:<br />
                    <input type="text" name="upduser" maxlength="30" value="<? echo $form->value("upduser"); ?>" /></td>
                                                          <td> Level:<br />
                                                          <select name="updlevel">
                                                              <option value="1">1 </option> //example settings
                                                              <option value="9">9 </option>
                                                            </select></td>
                                                          <td><br />
                                                            <input type="hidden" name="subupdlevel" value="1" />
                                                            <input type="submit" value="Update Level" /></td>
                                                        </tr>
                                                      </form>
                                                    </table></td>
                                                </tr>

Much help would be appreciated.

Mysqldump From Php File Not Populating With Data

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Any help on this would be greatly appreciated. I am trying to run a mysqldump from my site for my client when they want to back up. They basically will jsut click a button to run the backup.

With my script below, the backup file is genrated, but there is no table data in the file. There are the headers in the file with the server IP, linux versin, etc..., but there is no data in the file. Does this look right? Thank you for helping me out.

Ryan

Code: [Select]
ini_set ("display_errors", "1");
error_reporting(E_ALL);

$dbhost = 'localhost';
$dbuser = 'databaseuser';
$dbpass = 'password';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'database';
mysql_select_db($dbname);

$tableName = 'lots';
$backupFile1 = '/home/stuff/wwwroot/stuff/appnew/backup/';
$backupFile = $backupFile1.$dbname . date("Y-m-d-H-i-s") . '.gz';
$command = "mysqldump --opt -h$dbhost -u$dbuser -p$dbpass $dbname | gzip > $backupFile";
system($command);
$result = mysql_query($command);

Fetching Selected Value From One Dropdown To Populate Another Dropdown

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Hi freaks,

I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database.

The code I'm using to populate dropdown:
Code: [Select]
<?php
$con = mysql_connect("localhost","root","");
if(!$con)
{
  die ('Can not connect to : '.mysql_error());
}
mysql_select_db("ims",$con);
$result=mysql_query("select cat_id,cat_name from category");
echo "<select name=cat>";
while($nt=mysql_fetch_array($result))
{
echo "<option value=$nt[cat_id]> $nt[cat_name] </option>";
}
echo "</select>";
        mysql_close($con);
  ?>

Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like
   $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]");

Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow.
Help on this will be highly appreiable .
(Tell me if I'm not clear with my question)

Php Help Needed - Saving A Value In The Url To The Cahce And Populating A Field

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Hi wondering if anybody here could help me work out some code...

What I'm trying to do is save a value from a URL to populate a field on a form, but then I also want the value to be saved to the users cache so that is available to other pages on the site. Then once returning to the page with the form the field will repopulate with the data from the cache. I hope that makes sense.

example of link:
www.awebsite.com/pagename?id=1234

I know I can use Code: [Select]
$_GET["id"]; to read the value in the URL and populate the field on the form.

What I don't know how to do is save the value to the cache and then load it from cache automatically if the page doesn't have the value in the link.

Any help would be appreciated.