If this code is verbatim, then I would say it's because your $.post()
call does not include a "submit" param, which is what sendsuggest.php
is checking for in order to process the submission.
- jason
On May 16, 7:59 pm, riscphree <[EMAIL PROTECTED]> wrote:
> I've got a suggestion form that inserts into a db table when they
> submit, my jquery is located in a /js folder, while this page is in /,
> and my db settings are in /includes/
>
> Here is my $.post
>
> $.post("../sendsuggest.php",
> { subject: subjectVal, message: messageVal,
> theusername: theUser },
> function(data){
> $("#sendSuggestion").slideUp("normal", function()
> {
>
> $("#sendSuggestion").before('<h1>Success</
> h1><p>Thank you for your comments!</
> p>');
> });
> }
> );
>
> sendsuggest.php is something like this
>
> if(isset($_POST['submit'])) {
> $theuser = $_POST['theusername'];
> $subject = $_POST['subject'];
> $message = $_POST['message'];
> include("includes/database.inc");
> connect_to_cms();
> $sql = mysql_query("INSERT INTO suggestions(user,message,subject)
> VALUES('$theuser', '$message', '$subject')");
>
> }
>
> On my page, the slideUp works and this line: $
> ("#sendSuggestion").before('<h1>Success</h1><p>Thank you for your
> comments!</p>');
>
> works as well, but I can't figure out why it isn't inserting into the
> db. Any hints?
