AJAX-PHP Combination: $.post Checkbox value to PHP

Question

I'm trying for days to solve this problem, but with no results so far...
All I want to do is to POST checkbox value to php when the user check or uncheck it.

form.php:

<?php
if (isset($_POST['box'])) {
echo $_POST['box'];
} ?>

Script:

<script>
$(document).ready(function() {
 $('.box').on('change', function(event) { 
  var checkbox = $(event.target);
  var isChecked = $(checkbox).is(':checked');
  $.post('/form.php', {id: checkbox.attr('id'), value: isChecked});
 });
}); 
</script> //Credit cymen

Form:

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

<input id="box" class="box" name="box" type="checkbox" value="YESSSSSS!" />
<input id="box" class="box" name="box" type="checkbox" value="YESSSSSS!" />

</form>

(*) I get nothing on $_POST['box']

but I have a lot of them. How can php know which one I refer to? — wizard
– wizard, Commented Feb 15, 2013 at 11:51
@wizard Generally speaking the name attribute of the input corresponds to the request parameter that contains its value when you post the form. — Anthony Grist
– Anthony Grist, Commented Feb 15, 2013 at 11:51
@wizard, the element id is never sent to the server, only its name and value. For what you're doing here, there's no need for an id because you're using the class name to select the checkboxes. — MrCode
– MrCode, Commented Feb 15, 2013 at 11:54

Quentin · Accepted Answer · 2013-02-15 11:47:58Z

3

Your JavaScript is submitting this data to PHP:

 {id: checkbox.attr('id'), value: isChecked}

You have two pieces of data there: "id" and "value". You don't have a piece of data called "box", so $_POST['box'] will not be populated.

answered Feb 15, 2013 at 11:47

Quentin

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2 Comments

wizard Over a year ago

Do you mean: {id: checkbox.attr('box'), value: isChecked}? try it - it doesn't work.

Quentin Over a year ago

@wizard — No. You don't have <input box="anything"> do you? And that would still submit id something and you are looking for box in the data. If you want to submit the id as $_POST['box'] then {box:....

Boogeroos · Accepted Answer · 2013-02-15 11:57:04Z

0

$_POST array that You sent to form.php is't look $_POST['box']. It looks like this:

array(2) {["id"]=>string(3) "box" ["value"]=> string(4) "true"}

So You can get to vars like thist $_POST['id'] or $_POST['value']. You're declaring vars in JS in this fragment: {id: checkbox.attr('id'), value: isChecked}.

It's good to check in PHP what You have in vars using var_dump, eg. var_dump($_POST).

Use Firebug plugin to FF, to debug AJAX.

answered Feb 15, 2013 at 11:57

Boogeroos

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Comments

MrCode · Accepted Answer · 2013-02-15 12:03:36Z

0

I would change to:

$.post('/form.php', {name: checkbox.attr('name'), checked: isChecked, value : checkbox.val()});

On the PHP side you will receive:

Array
(
    [name] => box
    [checked] => true
    [value] => YESSSSSS!
)

You can access the POST data like:

echo $_POST['name']; // box
echo $_POST['checked']; // true/false
echo $_POST['value']; // YESSSSSS!

It's important that you include the checkbox value in the POST request because that is your only way of identifying which checkbox was changed.

answered Feb 15, 2013 at 12:03

MrCode

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1 Comment

wizard Over a year ago

Unfortunately it still not working....though everything seems in the right place.

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