javascript / php - modify input fields according to selection change

Question

javascript / php - modify input fields according to selection change

Can anyone help me with an issue I'm facing? I want to update multiple textfields whenever a new option is selected from my dropdown menu. I've written the following code, but it's not working as expected. Can someone figure out what's wrong? Thanks in advance.

        <script type="text/javascript">
            $("#combo").change(function() {
            var value = $(this).val();
            $.get(document, {type: value});
            });
        </script>


        <td class="bold"><label for='addType'>Choose Type to Add : </label>
            <select name="addType" id="combo">
                <option value="book">Book</option>
                <option value="author">Author</option>
                <option value ="publisher">Publisher</option>
                <option value="customer">Customer</option>
            </select>
        <?php
            $type['type'] = $_GET['type'];
            if ($type['type'] == 'author') {
                print "test";
            }
        ?>

php javascript ajax jquery

Answer 1

Answer №1

Here's an example of how the $.get function can be used:

 $.get("php_page.php", 
       {type: value},
       function(data) {
             $("#response").html(data)
      );

Add a div with the id "response" after your </select> tag to display the returned value:

 <div id="response">
 </div>

To handle your $.get AJAX request, create a separate page in the same directory named "php_page.php":

<?php
      echo $_GET['type'];
?>

If you're unfamiliar, PHP is a server-side language which means it completes its tasks before the page finishes loading. To utilize PHP functionalities after the page loads, AJAX comes into play.

Answer 2

Here's an example of how the $.get function can be used:

 $.get("php_page.php", 
       {type: value},
       function(data) {
             $("#response").html(data)
      );

Add a div with the id "response" after your </select> tag to display the returned value:

 <div id="response">
 </div>

To handle your $.get AJAX request, create a separate page in the same directory named "php_page.php":

<?php
      echo $_GET['type'];
?>

If you're unfamiliar, PHP is a server-side language which means it completes its tasks before the page finishes loading. To utilize PHP functionalities after the page loads, AJAX comes into play.

Answer 3

Answer №2

You are approaching this in an entirely incorrect manner!! I strongly recommend consulting the documentation on ajax within jQuery.

To help you get started with your project:

It is important to differentiate between the basic structure and the response you wish to receive from your php file. Please refer to this DEMO.

The HTML portion includes:

<select name="addType" id="combo">
    <option value="book">Book</option>
    <option value="author">Author</option>
    <option value ="publisher">Publisher</option>
    <option value="customer">Customer</option>
</select>
<div>Response: <span id="selection"></span></div>

While PHP can populate this, please note that it is not the actual response!

The JavaScript code is as follows:

$("#combo").change(function() {
    $.ajax({
        url: "/echo/html/",
        type: "POST",
        data: {"html":"Yeah: "+$(this).val()+" is cool!"},
        success: function(data) {
            $("#selection").html(data);
        }
    });
});

It would be ideal if you place this in a separate file.

Lastly, the "PHP file" identified by the url /echo/html has the following content:

<?php
  echo $_POST["html"];
?>

I hope this information proves beneficial.

Answer 4

You are approaching this in an entirely incorrect manner!! I strongly recommend consulting the documentation on ajax within jQuery.

To help you get started with your project:

It is important to differentiate between the basic structure and the response you wish to receive from your php file. Please refer to this DEMO.

The HTML portion includes:

<select name="addType" id="combo">
    <option value="book">Book</option>
    <option value="author">Author</option>
    <option value ="publisher">Publisher</option>
    <option value="customer">Customer</option>
</select>
<div>Response: <span id="selection"></span></div>

While PHP can populate this, please note that it is not the actual response!

The JavaScript code is as follows:

$("#combo").change(function() {
    $.ajax({
        url: "/echo/html/",
        type: "POST",
        data: {"html":"Yeah: "+$(this).val()+" is cool!"},
        success: function(data) {
            $("#selection").html(data);
        }
    });
});

It would be ideal if you place this in a separate file.

Lastly, the "PHP file" identified by the url /echo/html has the following content:

<?php
  echo $_POST["html"];
?>

I hope this information proves beneficial.

Answer 5

Answer №3

At first, let's talk about this specific section:

<?php
    $genre['genre'] = $_GET['genre'];
    if ($genre['genre'] == 'author') {
        print "demo";
}?>

This section will only be displayed when the page is initially loaded and only if genre=xxx is included in the URL query string. Keep in mind that PHP cannot be executed on the actual page once it has been loaded since the code is on the server-side.

For the jQuery component, give this a try:

$(document).ready(
    function(){
        $('#dropdown').change(function(){executeAjax();});
    }
);
function executeAjax(){
    $.get(
        'url', //I believe jQuery defaults to the current file if no URL is provided, but you can also use <?php echo $_SERVER['PHP_SELF'];?>
        {'genre' : $('#dropdown').val()},
        function(data){
                // What action should the AJAX call perform with the response data?
        },
        "html"
    );
}

In addition, it is recommended to create a div element using .html() to directly insert the returned data into it. This approach is much simpler than attempting to append the response after the last element above it.

Answer 6

At first, let's talk about this specific section:

<?php
    $genre['genre'] = $_GET['genre'];
    if ($genre['genre'] == 'author') {
        print "demo";
}?>

This section will only be displayed when the page is initially loaded and only if genre=xxx is included in the URL query string. Keep in mind that PHP cannot be executed on the actual page once it has been loaded since the code is on the server-side.

For the jQuery component, give this a try:

$(document).ready(
    function(){
        $('#dropdown').change(function(){executeAjax();});
    }
);
function executeAjax(){
    $.get(
        'url', //I believe jQuery defaults to the current file if no URL is provided, but you can also use <?php echo $_SERVER['PHP_SELF'];?>
        {'genre' : $('#dropdown').val()},
        function(data){
                // What action should the AJAX call perform with the response data?
        },
        "html"
    );
}

In addition, it is recommended to create a div element using .html() to directly insert the returned data into it. This approach is much simpler than attempting to append the response after the last element above it.

javascript / php - modify input fields according to selection change

Answer №1

Answer №2

Answer №3

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