why is my form not submiting using ajax

Question

bit of a selfish question but I am really strugerling with ajax in general so I need some help. So basically Im trying to update and sql database using an 'onblur' function. heres my code:

code on index.php

   function saveStatus(){
var status =document.getElementById("statusForm").value;
$.ajax({
                url: 'saveStatus.php',
                type: 'post',
                data: 'feed_id=' + status,
                success: function(result){
                }
              }    
    <form id="statusUpdate" action = "whosout.php" method="post"> 
        <input type="text" id="statusForm" onblur="saveStatus()" 
        placeholder="<?php if($status!=null){ echo '&lsquo;'.$status.'&rsquo;';}
        else { echo 'Enter your status here.';}?>"> 
        </form>

and code on saveStatus.php

    <?
     require 'core.php';
     require 'connect.php';

      $status = $_POST['feed_id'];
      $idPerson = $_SESSION['user_id'];
      $query = "UPDATE person SET status = '".mysql_real_escape_string($status)."' 
     WHERE idPerson = '$idPerson'"; 
    $query_run = mysql_query($query);
     ?>

at the moment the sql database does not update when i click of the input box. Any help would be great!!

Please attempt to narrow down error.

Patrick Geyer
– Patrick Geyer

2014-04-25 22:12:28 +00:00
Commented Apr 25, 2014 at 22:12 — Patrick Geyer
– Patrick Geyer, Commented Apr 25, 2014 at 22:12
Check for errors in your browser's console.

The Alpha
– The Alpha

2014-04-25 22:14:52 +00:00
Commented Apr 25, 2014 at 22:14 — The Alpha
– The Alpha, Commented Apr 25, 2014 at 22:14

Sharikov Vladislav · Accepted Answer · 2014-04-25 22:22:45Z

5

Answer:

You have to devide scripts and html output
You forget );.You have to close $.ajax block.
You miss } closing function saveStatus().

Code:

<script>
function saveStatus(){
    var status =document.getElementById("statusForm").value;
    $.ajax({
        url: 'saveStatus.php',
        type: 'post',
        data: 'feed_id=' + status,
        success: function(result){

        }
    }); // <-- Also, you forget `);` here 
} // <-- Also, you foget closing `}` here
</script>

<form id="statusUpdate" action = "whosout.php" method="post"> 
<input type="text" id="statusForm" onblur="saveStatus()" 
placeholder="<?php if($status!=null){ echo '&lsquo;'.$status.'&rsquo;';}
else { echo 'Enter your status here.';}?>"> 
</form>

Not error, but advice:

Also, note this: you used jQuery ajax function, so use jQuery in other cases too (you used pure JavaScript instead).

You getting value this way:

var status = document.getElementById("statusForm").value;

Use jQuery syntax instead:

var status = $("#statusForm").val();

Additional info

As, @Utkanos noticed in comments to this answer:

...all of which would be obvious if you look in the error console.

You have to use some debug tool, like FireBug or DevTools (Chrome, also in Chrome you can use CTRL+SHIFT+J). Not have to. You MUST do it.

edited Apr 25, 2014 at 22:22

answered Apr 25, 2014 at 22:13

Sharikov Vladislav

7,2999 gold badges53 silver badges93 bronze badges

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

2 Comments

Mitya Over a year ago

...all of which would be obvious if you look in the error console.

The Alpha Over a year ago

Also, mention about $('#statusForm').val() instead of var status =document.getElementById("statusForm").value;.

Collectives™ on Stack Overflow

why is my form not submiting using ajax

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related