How to call PHP function in dropdown menu?

Question

I want to call php function in dropdown menu by onchange event. I want with choose one of the options, the appropriate valuse are read from database and are list in another dropdown menu.

code:

<?php

function read() {

mysql_connect("localhost", "username", "password");

mysql_select_db("database_name");

$sql = mysql_query("SELECT name FROM table");

if (mysql_num_rows($sql)) {
    $select = '<select name="select">';
    while ($rs = mysql_fetch_array($sql)) {
        $select.='<option value="' . '">' . $rs['name'] . '</option>';
    }
}
$select.='</select>';
echo $select;
} 
?>

    <!--html code -->

    <select onchange="document.write('<?php read(); ?>');">
    <option value="0">a</option>
    <option value="1">b</option>
    <option value="2">c</option>
    </select>

This code output:

My desired output:

How can I get My desired output ? Thanks

You have to interact with the dropdown using Javascript. PHP does all its work before the page is shown and will not work on a rendered page without the help of a front-end scripting language, such as Javascript. — Wes Foster
– Wes Foster, Commented Jun 9, 2016 at 16:33
PHP runs on the server, and cannot be "called" by JS code. you can do an ajax request to invoke a php script, but that's not "calling" a php function. — Marc B
– Marc B, Commented Jun 9, 2016 at 16:39
Do you need a value from the (a,b,c) select list in order to display the new select list? I don't see that in your example. — bloodyKnuckles
– bloodyKnuckles, Commented Jun 9, 2016 at 16:46

CarlosCarucce · Accepted Answer · 2016-06-09 16:55:12Z

Just to explain:

PHP code is executed before the page is rendered in your user's browser (Server side).

In the other hand, Javascript is executed in the Client-side. It means that php finnished execution already.

If you wanna call a php function, you will have to make another request to the Server.

To do that "on the fly", you will have to use AJAX, as meantioned by @Jon in the comments.

Here is an example using jQuery (Just a javascript library, to simplify our task):

<script src="https://code.jquery.com/jquery-1.12.4.min.js"></script>
<script type="text/javasript">

    //Listen to select 'onchange' event
    $('select#your_select_id').change(function(){

        //Read selected value
        var inputValue = $(this).val();

        //Make an ajax call
        $.post('ajaxscript.php', { value: inputValue }, function(data){

            //The return of 'read' function will be accessible trough 'data'

            //You may create DOM elements here
            alert('Finnished');

        });

    });

</script>

and here is our ajaxscript.php content:

<?php
//Declare (or include) our function here
//function read(){ ... }

$value = $_POST['value']; //Selected option
//...

echo read();

Passionate Coder · Accepted Answer · 2016-06-09 17:18:45Z

1

Hi You can also use javascript form submit in this and call a php function

<?php
function read() {
mysql_connect("localhost", "username", "password");
mysql_select_db("database_name");
 $sql = mysql_query("SELECT name FROM table");
if (mysql_num_rows($sql)) {
$select = '<select name="select">';
while ($rs = mysql_fetch_array($sql)) {
$select.='<option value="' . '">' . $rs['name'] . '</option>';
 }
}
 $select.='</select>';
echo $select;
 } 
 if (isset($_POST['value'])) {
  read($_POST['value']);
}
    ?>
  <form method="POST">
 <select name="value" onchange="this.form.submit()">
 <option value="0">a</option>
<option value="1">b</option>
<option value="2">c</option>
</select>
 </form>

answered Jun 9, 2016 at 17:18

Passionate Coder

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1 Comment

ParisaN Over a year ago

I test this and worked but Is it logically true (send form) ?

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