PHP - Dropdown Not Showing Up For Every Row In Database.

Hi, I'm trying to get data from one field in a table (database). But I get undesirable result: Here is my code ->

<?php
$result2 = mysql_query("SELECT DISTINCT theme FROM  mytable ")
               or die(mysql_error());
         
            
            while($row2 = mysql_fetch_array( $result2 ))
               {
?>
                  <form method="post" action='<?php echo $_SERVER["PHP_SELF"]; ?>'>

                     <select name='themes'">
<?php
                   
                  $arr= array($row2['theme']);
      
                  foreach($row2 as $value)
                  {
                     echo "<option value='$value'><b>". $value."</b> </option><br> ";
                     
                  }
                  
               }
?>

The attached image file show the result that I don't wont. (It's not a dropdown).
Is there anyone who may help me, I spent a lot of time to find out but I can't.

Thanks a lot for your help

mysql.php
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ckeditor",$con);
?>
---------------------------------------------------------
add.php
<?php
include("mysql.php");


if(isset($_POST["button2"]))
{
$sql="INSERT INTO cktext (section)
VALUES
('$_POST[select2]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }

?>

-----------------------------------------------------------------
home.php

<form id="form1" name="form1" method="post" action="add.php">
<tr>
    <td>Section:</td>
    <td><select name="select2" id="select2">
     <option selected="selected" value="MALE">Male</option>
  <option selected="" value="FEMAIL">FEMAIL</option>
    </select></td>
  </tr>
<input type="submit" name="button2" id="button2" value="Upload" />
  </form>

In DATABASE :- cktext table attribute "section" is varchar type.

BUT IT RETURN ME BLANK OUTPUT.

Pls anyone tell me to connect database

hi. I am stuck on the dropdown box value calculation. anyone can help me on this problem ? I appreciate it.

I have two dropdown boxes on my webpage, and the values that in the dropdown boxes are retrived from the mysql database.

dropdown box "A" displays two differnt values, which are "Chassise_Name" and "Chassise_price", and The dropdown box "B" displays the quantity of the item.

    Dropdown box "A"                                 Dropdown box "B"
 SuperServer 7036A-T(Black) $90                       3


"Superserver 7036A-T(Black)" is retrieved from the Chassis_Name ;
"$90" is retrieved from the chassis_Price;
"3" is retrieved from the Quantity in other table.


Question:

how can I take only the price on the Dropdown box "A" multiply the number in the Dropdown box "B"?

Here is my code:



Code: [Select]
echo "<table border='1'>";
echo "<tr><td>";

$result = mysql_query("SELECT Chassis_Name, Chassis_Price FROM Chassis where Chassis_Form_Factor='ATX'")
or die(mysql_error()); 
echo '<select name="Chassis" onChange="change()">';
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
// Print out the contents of each row into a table
echo '<option Name="Chassis">',$row['Chassis_Name'],' &nbsp;&nbsp;&nbsp;',$row['Chassis_Price'],'&nbsp;','</option>';
}
echo '</select>';

echo "</td>";






echo "<td>";

$result= mysql_query("SELECT Number FROM Quantity")
or die(mysql_error()); 
echo '<select name="Quantity" Value="Quantity" onChange="change()" >';

// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result)) {
// Print out the contents of each row into a table

echo '<option Name="Quantity">',$row['Number'],'</option>';
}
echo '</select>';




echo "</td></tr>";
echo "</table>";


?>

Hi,

I have a MySQL database called "2011_database" that has a table called "2011_list." In that table I have fields, amongst others, called "name" and "district."

I need to find way to get the data from the table and put them into a drop down list on other PHP page. But they need to be listed as "name - district" on one line.

I am PHP beginner and if I understand it correctly there need to be two references to get all the data in all records, a third reference to merge them together with " - " in between; and what eludes me the most, putting them in a drop down menu.

Any help is greatly appreciated
Thanks

Hey Everyone...

First off, I am only a young web developer and i'm working on a school project and am making a text-based game online...

Now what i'm having trouble with...

I want a drop-down list that has a list of characters classes
Clubber
Mixer
Sauceror
Tamer

And I want whatever is selected to be placed into the database along with the username/password (THIS ALL WORKS FINE JUST NOT THE DROP DOWN LIST)

All help appreciated

I am very new new php (wrote my first PHP script 5 Days ago) and am trying to give myself a crash course but I have hit a pit-stop which is killing me a little!

I hope that title makes sense.... Basically I created PHP script to take data from a database and display in, I then wrote some code to use a drop down menu to order that data. That all worked ok until I tried to utilise some pagination. I can make the pagination work, and I can make the ordering work, but not at the same time! At the moment the code that I have will allow me to order the list and almost paginate it. There are 40 results and I want to display 10 at a time. When not using the ordering code I can paginate it perfectly but when I try to intergrate the two bits of code it will only display the first 10 results and not give me an option to go to the next page to see the rest!

The code:

if (!isset($_GET['start'])) {

$_GET['start'] = 0; }

$per_page = 10; $start = $_GET['start'];
if (!$start)
   $start = 0;
 
 
 
$sort = @$_POST['order'];   
if (!empty($sort)) {   
$get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn
            FROM booktable
            ORDER BY ".mysql_real_escape_string($_POST['order'])." ASC
            LIMIT $start, $per_page");
}
 else { 
$get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn 
                FROM booktable 
                ORDER BY bookname ASC
                LIMIT $start, $per_page"); 
}
 
$record_count = mysql_num_rows($get);
 
 
?>
<?php
if (isset($_GET['showerror']))
$errorcode = $_GET['showerror'];
else
$errorcode = 0;
?>

 wont include all the html rubbish and the ordering menu!
<div id="mid">
 
<?php
echo "<table>";
echo "<tr>";
echo "<th>";
 
echo "</th>";
echo "<th>";
echo "Book Title";
echo "</th>";
echo "<th>";
echo "Book Author";
echo "</th>";
echo "<th>";
echo "Book Publisher";
echo "</th>";
echo "<th>";
echo "Book ISBN";
echo "</th>";
echo "<th>";
 
echo "</th>";
echo "</tr>";
while ($row = mysql_fetch_assoc($get))
{
 // get data
 $bookname = $row['bookname'];
 $bookauthor = $row['bookauthor'];
 $bookpub = $row['bookpub'];
 $bookisbn = $row['bookisbn'];
 
 
 
    echo "<tr>";
    echo "<td>";
    echo "<a href='addtolist.php?bookname=".$bookname."&bookauthor=".$bookauthor."&bookpub=".$bookpub."&bookisbn=".$bookisbn."'>Add to basket</a>";
 
    echo "</td>";
 
    echo "<td>";
    echo $bookname;
    echo "</td>";
 
    echo "<td>";
    echo $bookauthor;
    echo "</td>";
 
    echo "<td>";
    echo $bookpub;
    echo "</td>";
 
    echo "<td>";
    echo $bookisbn;
    echo "</td>";
 
    echo "</tr>";
 
 
}
 
echo "</table>";
 
$prev = $start - $per_page;
$next = $start + $per_page;
 
if (!($start<=0))
       echo "<a href='products.php?start=$prev'>Prev</a> ";
 
       //set variable for first page number
$i=1;
 
//show page numbers
for ($x = 0; $x < $record_count; $x = $x + $per_page)
{
 if ($start != $x)
    echo "<a class='pagin' href='products.php?start=$x'> $i </a>";
 else
    echo "<a class='pagin' href='products.php?start=$x'><b> $i </b></a>";
 $i++;
}
 
//show next button
if (!($start >= $record_count - $per_page))
       echo "<a class='pagin' href='products.php?start=$next'> Next </a>";
?>


Thank you so much for reading!

This is actually a page to edit the drama details.
Database-  3 tables

1. drama                           dramaID    drama_title
                                              1              friends

2. drama_genre             drama_genreID     dramaID    genreID
                                               1                        1               2
                                               2                        1               1
                                               3                        1               3
3. genre                              genreID       genre
                                                1            comedy
                                                2            romance
                                                3            family
                                                4            suspense
                                                5            war
                                                6            horror

<?php       

$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'drama';
$link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');

$id = $_GET['dramaID'];

$link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');
$sql2 = "SELECT * from drama , drama_genre , genre
WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID";
$status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2));

$link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql');
$sql3 = "SELECT * from genre ";
$status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3));

<form method='Post' action='doUpdate.php' enctype="multipart/form-data">

<?php while ($row2 = mysqli_fetch_assoc($status2)) { ?>
                  <td height="1"></td>
               
                             <td><select name="gdrop">      
                     
      <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option>
                     
      <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?>
                        
      <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option>
                     
                     
      <?php }   mysqli_close($link3);  ?>
                     
 <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo          $row2['drama_genreID']; ?>"/>
                 </select>

<input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/>
<input type="submit" Value="Update"/>
 
The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama.

I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved.

And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection.


                                       

Hello,

If i have a title like "It's Wonderful!" it always gets displayed as "It?s Wonderful!" and the ? is always in a black diamond.

How can i get it to show a ' properly?

Thanks

I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database.  I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window.  I do not know how to correct this.  I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page.  Suggestions on what I need to look for to correct would be great.  I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. 
Thank you for any assistance I may get!!  - Amy -
 
Code: [Select]
<body><font color="006600">
 <div style="background-color:#f9f9dd;">
  <fieldset>
<h1>Asset Entry Results</h1>
<?php
  // create short variable names
  $tag=$_POST['tag'];
  $serial=$_POST['serial'];
  $category=$_POST['category'];
  $status=$_POST['status'];
  $branch=$_POST['branch'];
  $comments=$_POST['comments'];

  if (!$tag || !$serial || !$category || !$status || !$branch) {
     echo "You have not entered all the required details.<br />"
          ."Please go back and try again.";
     exit;
  }

  if (!get_magic_quotes_gpc()) {
    $tag = addslashes($tag);
    $serial = addslashes($serial);
    $category = addslashes($category);
    $status = addslashes($status);
    $branch = addslashes($branch);
    $comments = addslashes($comments);
  }

  //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory');
  $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database");

  $query = "insert into assets values
            ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')";
  $result = $db->query($query);

  if ($result) {
      echo  $db->affected_rows." asset inserted into Inventory.";
  } else {
     echo "An error has occurred.  The item was not added.";
  }

  $db->close();
?>
  </fieldset>
 </div>
</body>

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)

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 a very tricky php problem here. At least for me.
On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered.

So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out.

Description of what I'm trying to do:
I'm generating the the boxes with a call to my database for the list of checkboxes.
Then I'm building an array with all the possible items that could be checked to compare against.
 
Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient"

       Code: [Select]
<div class="text">Job Type&nbsp;(<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div>
<div class="field">

    <?PHP

    $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database.");
    $db=mysql_select_db ("database", $connection) or die (mysql_error());
    $query = "SELECT type FROM typelist ORDER BY type ASC";
    $sql_result = mysql_query($query, $connection) or die (mysql_error());
    $i=1;
    echo '<table valign="top"><tr><td>';
    while ($row = mysql_fetch_array($sql_result)) {
    $type = $row["type"];
        if ($i > 1 && $i % 26 == 0)
        echo '</td><td>';
        else if ($i)
        echo '';
        ++$i;
       
        $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)");
       
        $dbTypes = explode(',',$row['type']);
       
        foreach ($aTypes as $type){
       
        if(in_array($aType,$dbTypes)){
       
      echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>";
       
        }else
       
        {
       
            echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>";  
       
               
        }

        }

    }
      echo '</td></tr></table>';
   
    ?>

</div>