jazzman1

I've checked your script and the form has been submitted in the way, I expected to be. Do you use any js debuger tools ?

@Doc, take a look at this link -> http://www.reg.ca/faq/PhpMyAdminTutorial.html

@Doc, I don't know why your database structure looks like this: Array ( o=> u_id [Field] => u_id [1] => int(11) [Type] => int(11) [2] => NO [Null] => NO [3] => PRI [Key] => PRI [4] => [Default] => [5] => auto_increment [Extra] => auto_increment ) Array ( o=> id [Field] => id [1] => int(11) [Type] => int(11) [2] => NO [Null] => NO [3] => [Key] => [4] => [Default] => [5] => [Extra] => ) Array ( o=> site_id [Field] => site_id [1] => int(11) [Type] => int(11) [2] => NO [Null] => NO [3] => [Key] => [4] => [Default] => [5] => [Extra] => ) Array ( etc..... ) I'd give you advise, to create the same table again..

Try this: print_r($row[0]['o'], true)

Change 0 to o in print_r function. print_r($row['o'], true)

Yes, now is Ok, but why you get multiple arrays in the same object, that's why you cannot insert the data from your form. Give me a result please. //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx')or die(mysql_error()); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx', true) or die(mysql_error()); $recruit_db = mysql_select_db('doggysty_recruitment', $recruit_link)or die(mysql_error()); $webspell_db = mysql_select_db('doggysty_site', $webspell_link)or die(mysql_error()); $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_assoc($result)) { echo '<pre>'.print_r($row[0], true).'</pre>'; } exit;

Maybe, I'm doing something wrong, put in print_r() in the while loop. //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx')or die(mysql_error()); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx', true) or die(mysql_error()); $recruit_db = mysql_select_db('doggysty_recruitment', $recruit_link)or die(mysql_error()); $webspell_db = mysql_select_db('doggysty_site', $webspell_link)or die(mysql_error()); $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_array($result)) { echo '<pre>'.print_r($row, true).'</pre>'; } exit;

The tests, do you make on the local machine or a server ?

You have a syntax error, copy/paste this: //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx')or die(mysql_error()); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxx', true) or die(mysql_error()); $recruit_db = mysql_select_db('doggysty_recruitment', $recruit_link)or die(mysql_error()); $webspell_db = mysql_select_db('doggysty_site', $webspell_link)or die(mysql_error()); $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_array($result)) { $arr = $row; } echo '<pre>'.print_r($arr, true).'</pre>'; exit;

Ok, that's correct. Now go to include.php file and copy/paste this code there: $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_array($result)) { $arr = $row; } echo '<pre>'.print_r($arr, true).'</pre>'; exit;

Hm.... it's very weird. Open up your phpmyadmin application and go to database named " doggysty_recruitment ", click on the SQL tab and copy/paste this code: EXPLAIN `doggysty_recruitment`.`applications`

Put in php die() function, to be like this: //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxx') or die(mysql_error()); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxxxx') or die(mysql_error()); $recruit_db = mysql_select_db('doggysty_recruitment', $recruit_link) or die(mysql_error()); $webspell_db = mysql_select_db('doggysty_site', $webspell_link) or die(mysql_error());

No, Doc, just assign them to new variables and try again, for example: //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxxxxxx'); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxxxxxx', true); $db_name1 = mysql_select_db('doggysty_recruitment', $recruit_link); $db_name2 = mysql_select_db('doggysty_site', $webspell_link);

@Doc, I think, I got it. According the file bellow, you're overriding or overwriting (I'm not sure how proper to say in English) mysql_select_db('name'), you must assign it to different variables. //connect to databases $recruit_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxxxxxx'); $webspell_link = mysql_connect('localhost', 'doggysty_recruit', 'xxxxxxxxxx', true); mysql_select_db('doggysty_recruitment', $recruit_link); mysql_select_db('doggysty_site', $webspell_link);

Your form fields and data has came from it is just fine. Do you have access to databases through any desktop or browser applications, like phpmyadmin, mysql workbench and etc ?

When you execute this code, it can be shown the full structure of table `applicaions` from db name `doggysty_recruitment`. There is something wrong, double check again. $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_array($result)) { $arr = $row; } echo '<pre>'.print_r($arr, true).'</pre>'; exit;

Compare, what you want to do. You have to create a new table structure according the html form fields or backward. // form data values [name] => Doc20 [i_name] => Doc [username] => Doc [steam] => N/A [xfire] => bagpiper4414 [email] => doc@test.com [games] => all the games I play [age] => 37 [location] => USA: North Coast [microphone] => 2 [why_join] => Because I like it! [clan_history] => Well, this is my first [recruit_style] => 1 [recruit] => Doggy Style [userIP] => x.x.x.x [time] => 1344143935 [submit] => Send Application // table structure Field] => notes [1] => longtext [Type] => longtext [2] => NO [Null] => NO [3] => [Key] => [4] => [Default] => [5] =>

Try this: <!DOCTYPE html> <html> <head> <link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" rel="stylesheet" type="text/css"/> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script> <script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script> <style type="text/css"> .draggable { width: 75px; height: 25px; background: silver; padding: 10px; margin-bottom: 2px; } #droppable { position: absolute; left: 250px; top: 0; width: 125px; height: 75px; background: gray; color: white; padding: 10px; } </style> <script> $(document).ready(function() { $("div.draggable").draggable(); $("#droppable").droppable({ drop: function() { alert('dropped'); } }); }); </script> </head> <body style="font-size:62.5%;"> <div id="droppable">Drop here</div> <div class="draggable">Drag me 1</div> <div class="draggable">Drag me 2</div> <div class="draggable">Drag me 3</div> <div class="draggable">Drag me 4</div> </body> </html>

Why do you remove all box ? $('.box').remove();

Are you sure that your computer or server supports php short tag ?

You must define a post method in your html form, by default it's a GET method. In your js script you use jquery post method. $.post('save_profile.php',$(this).serialize(),function(data)

Add this instead of code above. echo '<pre>'.print_r($site_id, true).'</pre>'; $result = mysql_query('EXPLAIN `doggysty_recruitment`.`applications`', $recruit_link) or die(mysql_error()); $arr = array(); while ($row = mysql_fetch_array($result)) { $arr = $row; } echo '<pre>'.print_r($arr, true).'</pre>'; exit;

Sorry, I meant to write -> site_id

If you want an event to work on your page, you should call it inside the $(document).ready() function.

Thanks