PHP - Javascript Select Dropdown

Hi,

I am doing an EDIT user page and would like to check records with a dropdown and then set as selected, please help!


<select name="BrokerID" class="small-input">

<option value="Please select an option">Please select a Broker</option> 

<?php 

while($row2 = mysql_fetch_array($broker))

{

echo '<option name="BrokerID" value="'.$row2['BrokerID'].'">'.$row2['BrokerName'].'</option>';

}

?>

</select>

I'm trying to figure out why the options aren't appearing inside the select dropdown. Any ideas why?

Code: [Select]
echo "<label for=" . $row2['fullName'] . ">" . $row2['fullName'] . "</label>";
                       echo "<select name=" . $row2['fullName'] . " id=" . $row2['fullName'] . " class=dropdown title=" . $row2['fullName'] . " />";
                       if ($styleID == 1 || $styleID == 2 || $styleID == 6) {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID = 3
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }   
                       } else {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID IN (1,2,6)
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }
                       }
                       echo "</select>";
                   }

Hi,  I'm a php newbie, with some mysql experience. I have a mysql database as follows:
Database=watch, Table=events - fields id, reportno, sdate, comments

What I need is:
1. A dropdown list to display reportno from mysql database.
2. Depending on which reportno I choose, I'd like to open a popup(or separate) page to display the stored information.

Tks in advance for any help

i am new to web designing,   i don't know much about javascript.   <select id="plan" align="center" valign="center">

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

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)

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

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>";
?>
 

 

I have 2 queries that I want to join together to make one row  

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>

I'm trying to have it shows the data from my array as each option will have a value of the user's id and the text for the option will be their name. As of right now it only shows the last person.


Code: [Select]

<?php echo form_label('Recipient', 'recipient'); ?>

<?php 
$data = array(
  'name'   => 'to',
  'class'  => 'required'
);
<?php echo form_label('Recipient', 'recipient'); ?>
                                    <?php 
                                    $data = array(
                                    'name'   => 'to',
                                    'class'  => 'required'
                                    );
                                    $options = array();
foreach($users AS $user)
{
    $options = array (
    $user->user_id => $user->first_name.' '.$user->last_name
    );     
}   
?>

<?php echo form_dropdown($data, $options); ?>

Hi all. I kinda need some from a more advanced PHP expert. I have a table that displays time slots that are available to be booked. As you can see I have a table with two columns, The 1st column displays the times and the 2nd column has a link that says 'Available'. I want to be able to put all this in a select dropdown box to save space, but how can I do this?

<?php 
	
	$doc = JFactory::getDocument();
	$doc->addStyleSheet(JURI::root(false)."components/com_pbbooking/user_view.css");
?>

<style>
table#pbbooking td, table#pbbooking th {padding: 0em;}
</style>


<h1><?php echo JText::_('COM_PBBOOKING_DAY_VIEW_HEADING').' '.Jhtml::_('date',$this->dateparam->format(DATE_ATOM),JText::_('COM_PBBOOKING_DAY_VIEW_DATE_FORMAT'));?></h1>
<table id="pbbooking">
<!-- Draw header row showing calendars across the top....-->
	<tr>
		<th></th> <!-- first column left blank to display time slots -->
		<?php foreach ($this->cals as $cal) :?>
			<th><?php echo $cal->name;?></th>
		<?php endforeach;?>
	</tr>


	<!-- draw table data rows -->

	<?php while ($this->day_dt_start <= $this->dt_last_slot) :?>
		<?php $slot_end = date_create($this->day_dt_start->format(DATE_ATOM),new DateTimezone(PBBOOKING_TIMEZONE));?>
		<?php $slot_end->modify('+ '.$this->config->time_increment.' minutes');?>
		<tr>
			<th><?php echo Jhtml::_('date',$this->day_dt_start->format(DATE_ATOM),JText::_('COM_PBBOOKING_SUCCESS_TIME_FORMAT'));?></th>
			<?php foreach ($this->cals as $cal) :?>
				<td class="pbbooking-<?php echo (!$cal->is_free_from_to($this->day_dt_start,$slot_end)) ? 'free' : 'busy';?>-cell">
					<?php if ($this->day_dt_start>date_create("now",new DateTimeZone(PBBOOKING_TIMEZONE)) && !$cal->is_free_from_to($this->day_dt_start,$slot_end)) :?>
						<a href="<?php echo JRoute::_('index.php?option=com_pbbooking&task=create&dtstart='.$this->day_dt_start->format('YmdHi').'&cal_id='.$cal->cal_id);?>">
							<?php echo (!$cal->is_free_from_to($this->day_dt_start,$slot_end)) ? JText::_('COM_PBBOOKING_FREE') : JText::_('COM_PBBOOKING_BUSY');?>
						</a>
					<?php else :?>
						<?php echo JText::_('COM_PBBOOKING_BUSY');?>
					<?php endif;?>
				</td>
			<?php endforeach;?>
		</tr>
		<?php $this->day_dt_start->modify('+ '.$this->config->time_increment.' minutes');?>
	<?php endwhile;?>

	<!-- end draw table data rows-->

</table>

I currently have a web page which the user selects the value from the drop menu.
I then want that value to go into another form in the same page inside a hidden field.
Which will then be submitted to the processing php.

The second option is the user selects the drop menu  then the processing php would have to grab
the value without a submit button because I am using a save button on the same page in the
other form.

Any help

Hi, I'm wondering how i would go about making a drop down menu in php to insert the option into users > Maint of my database, if anyone has any tutorials or examples it would be great, Thanks a lot.