dynamic drop down box?

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I got a database table called category as shown:

$\"enter$

I am trying to do a d

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野趣味

2020-11-22 07:28
Here is an example that will do what you want. Essentially, you can use jQuery / AJAX to accomplish this.

I updated my example code to match your server login / table / field names, so if you copy/paste these two examples into files (call them tester.php and another_php_file.php) then you should have a fully working example to play with.

I modified my example below to create a second drop-down box, populated with the values found. If you follow the logic line by line, you will see it is actually quite simple. I left in several commented lines that, if uncommented (one at a time) will show you what the script is doing at each stage.

FILE 1 -- TESTER.PHP
```
<html>
    <head>
        <script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
        <script type="text/javascript">
            $(function() {
//alert('Document is ready');

                $('#stSelect').change(function() {
                    var sel_stud = $(this).val();
//alert('You picked: ' + sel_stud);

                    $.ajax({
                        type: "POST",
                        url: "another_php_file.php",
                        data: 'theOption=' + sel_stud,
                        success: function(whatigot) {
//alert('Server-side response: ' + whatigot);
                            $('#LaDIV').html(whatigot);
                            $('#theButton').click(function() {
                                alert('You clicked the button');
                            });
                        } //END success fn
                    }); //END $.ajax
                }); //END dropdown change event
            }); //END document.ready
        </script>
    </head>
<body>

    <select name="students" id="stSelect">
        <option value="">Please Select</option>
        <option value="John">John Doe</option>
        <option value="Mike">Mike Williams</option>
        <option value="Chris">Chris Edwards</option>
    </select>
    <div id="LaDIV"></div>

</body>
</html>
```
FILE 2 - another_php_file.php
```
<?php

//Login to database (usually this is stored in a separate php file and included in each file where required)
    $server = 'localhost'; //localhost is the usual name of the server if apache/Linux.
    $login = 'root';
    $pword = '';
    $dbname = 'test';
    mysql_connect($server,$login,$pword) or die($connect_error); //or die(mysql_error());
    mysql_select_db($dbname) or die($connect_error);

//Get value posted in by ajax
    $selStudent = $_POST['theOption'];
    //die('You sent: ' . $selStudent);

//Run DB query
    $query = "SELECT * FROM `category` WHERE `master` = 0";
    $result = mysql_query($query) or die('Fn another_php_file.php ERROR: ' . mysql_error());
    $num_rows_returned = mysql_num_rows($result);
    //die('Query returned ' . $num_rows_returned . ' rows.');

//Prepare response html markup
    $r = '  
            <h1>Found in Database:</h1>
            <select>
    ';

//Parse mysql results and create response string. Response can be an html table, a full page, or just a few characters
    if ($num_rows_returned > 0) {
        while ($row = mysql_fetch_assoc($result)) {
            $r = $r . '<option value="' .$row['id']. '">' . $row['name'] . '</option>';
        }
    } else {
        $r = '<p>No student by that name on staff</p>';
    }

//Add this extra button for fun
    $r = $r . '</select><button id="theButton">Click Me</button>';

//The response echoed below will be inserted into the 
    echo $r;
```
To answer your question in the comment: "How do you make the 2nd drop down box populate fields that are only relevant to a selected option from the 1st drop down box?"

A. Inside the .change event for the first dropdown, you read the value of the first dropdown box:

$('#dropdown_id').change(function() { var dd1 = $('#dropdown_id').val(); }

B. In your AJAX code for the above .change() event, you include that variable in the data you are sending to the 2nd .PHP file (in our case, "another_php_file.php")

C. You use that passed-in variable in your mysql query, thereby limiting your results. These results are then passed back to the AJAX function and you can access them in the success: portion of the AJAX function

D. In that success function, you inject code into the DOM with the revised SELECT values.

That is what I am doing in the example posted above:
1. The user chooses a student name, which fires the jQuery .change() selector
2. Here is the line where it grabs the option selected by the user:
  
  var sel_stud = $(this).val();
3. This value is sent to another_php_file.php, via this line of the AJAX code:
  
  data: 'theOption=' + sel_stud,
4. The receiving file another_php_file.php receives the user's selection here:
  
  $selStudent = $_POST['theOption'];
5. Var $selStudent (the user's selection posted in via AJAX) is used in the mysql search:
  
  $query = " SELECT * FROM `category` WHERE `master` = 0 AND `name` = '$selStudent' ";
  
  (When changing the example to suit your database, the reference to $selStudent was removed. But this (here, above) is how you would use it).
6. We now build a new <SELECT> code block, storing the HTML in a variable called $r. When the HTML is fully built, I return the customized code back to the AJAX routine simply by echoing it back:
  
  echo $r;
7. The received data (the customized <SELECT> code block) is available to us inside the AJAX success: function() {//code block}, and I can inject it into the DOM here:
  
  $('#LaDIV').html(whatigot);
And voila, you now see a second dropdown control customized with values specific to the choice from the first dropdown control.

Works like a non-Microsoft browser.
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