CODEWITHSUNDEEP
phpjquerymysqlajax
Jonathon Doesen
Jonathon Doesen

Reputation: 563

How to submit form via ajax with jQuery?

I need to submit an html form via .ajax with jQuery. When I submit it normally, not via ajax, everything works great. It is in switching to ajax that I encounter problems.

The form has changing number of inputs. (One each for each user in a particular group, groups can have a differing number of users). So I name the inputs like so:

<input type="hidden" name="<?php echo 'sampleA['. $i .']'; ?>" value="sampleValue" />

Where $i is a counter so the inputs are sampleA['1'] and sampleA['2'] and so on.

I have another input for each user that is done the same way, which we can call sample2['1'] and so forth.

So with php I just get these 2 arrays and run through them with a similar counter and insert them to my db as need. This all works well. But I need to make it ajax.

How do I pass those two arrays via jQuery's .ajax function to a php script, and then how do I handle it in php?

I've tried using .serialize on my form, but that serializes both inputs on all the users into one array and I have been unable to de-serialzie that correctly on the php side of things.

The following doesn't work, but I will include it here to show you what I've been doing:

function updateOnCourt() {             
    var thearray = $('#sub-form').serialize();
    $.ajax({
      url: 'sub-update.php',
      method: 'POST',
      data: thearray,
      success: function() {
        alert('Success: '+thearray);
      }
    });
  }

and then in my PHP script:

<?php 
$size = count($_POST['thearray']);
$finally = json_decode($_POST['thearray'], true);

// start a loop in order to update each record
$i = 1;

while ($i < ($size+1)) {


// define each variable
$pid = $finally[$i];
$playing = $finally[$i];


$query = "UPDATE players SET 
    on_court = '{$playing}'
WHERE id = {$pid}";

mysql_query($query, $connection);

$i = $i + 1;

} ?>

What am I doing wrong?

Upvotes: 1

Views: 231

Answers (1)

charlietfl
charlietfl

Reputation: 171700

thearray is only a variable name in javascript, so there is no $_POST['thearray'] and the data is not sent as JSON so no need to use json_decode

You want to access the name attributes in the form controls:

Try looking at $_POST['sampleA'].

Look at docs for serialize() will see that it sends the data in same format as if form was submitted without AJAX

API Reference: http://api.jquery.com/serialize/

Upvotes: 1

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