Passing variables from JQuery to PHP using $.post

Question

I have a form (#searchform) that I want to submit multiple times without reloading the page, so I use jQuery and the post-method. I want to store which checkboxes are checked in a variable named target (array, list or whatever) and which radio button is selected in the variable diff.

The variables also need to be available in PHP. I tried passing them as ftar and fdiff in so many ways, but nothing worked.

Any help is much appreciated!

JQuery in submit.js:

$(document).ready(function () {
    var target = new Array();
    $('#searchform input:checked').each(function() {
        target.push($(this).attr('value'));
    });

    var diff = $('#searchform').find("input[class='diff']").val();

    $.post('index.php', {ftar: target, fdiff: diff});
});

HTML and PHP in index.php:

<!DOCTYPE html>
<html lang="en-US">
<head>
    <title></title>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
    <script src="js/submit.js"></script>
</head>

<body>

<?php 

    ini_set('display_errors',1); 
    error_reporting(E_ALL);
    $link = new mysqli("localhost", "root", "password", "mytable");

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }

    $target = $_POST['ftar'];
    $diff = $_POST['fdiff'];
?>

<div id="right">
    <div class="content">
        <form id="searchform" method="POST" action="/">
            <input type="checkbox" value="Check1">
            <input type="checkbox" value="Check2">
            <input type="checkbox" value="Check3">

            <input type="radio" class="diff" value="Radio1">
            <input type="radio" class="diff" value="Radio2">
        </form>
    </div>
</div>

</body>
</html>

You aren't doing anything with the $target and $diff variables. HOw are you checking that there is an issue? — dannrob
– dannrob, Commented Jun 14, 2013 at 17:00
try to build a valid HTML first and validate it. Then you can go to the javascript part — Elzo Valugi
– Elzo Valugi, Commented Jun 18, 2013 at 12:34

U.P · Accepted Answer · 2013-06-14 17:01:50Z

1

You should use

JSON.stringify

$.post('index.php', JSON.stringify({ftar: target, fdiff: diff}));

answered Jun 14, 2013 at 17:01

U.P

7,4527 gold badges41 silver badges63 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Valdo Raya · Accepted Answer · 2013-06-14 17:03:56Z

Did you tried to use $(#searchform).serialize();?

$.post('index.php', $(#searchform).serialize());

You should also add a name to your fields in the form

<form id="searchform" method="POST" action="/">
        <input name="check[]" type="checkbox" value="Check1">
        <input name="check[]" type="checkbox" value="Check2">
        <input name="check[]" type="checkbox" value="Check3">

        <input name="radio[]" type="radio" class="diff" value="Radio1">
        <input name="radio[]" type="radio" class="diff" value="Radio2">
    </form>

Fetch field as an array (only marked fields will be passed)

    $target = $_POST['check'];
print_r($target)
$target = $_POST['radio'];
print_r($target)`

I did try serialize, but find it unneccessarily complicated to retrieve informations about which checkboxes are checked from the string that is returned. Didn't solve the post problem either.

icebreaker · Accepted Answer · 2013-06-18 12:27:17Z

0

You should add name="ftar" and name="fdiff" to according html inputs. Then serialize, JSON.stringify() or just submit.

answered Jun 18, 2013 at 12:27

icebreaker

1,27418 silver badges34 bronze badges

Collectives™ on Stack Overflow

Passing variables from JQuery to PHP using $.post

3 Answers 3

JSON.stringify

Comments

1 Comment

Comments

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

JSON.stringify

Comments

1 Comment

Comments

Related