jQuery ajax call won't update mysql after pressing back button - php

Question

I have a form that uses ajax to submit data to a mysql database, then sends the form on to PayPal.
However, after submitting, if I click the back button on my browser, change some fields, and then submit the form again, the mysql data isn't updated, nor is a new entry created.
Here's my Jquery:
$j(".submit").click(function() {
var hasError = false;
var order_id = $j('input[name="custom"]').val();
var order_amount = $j('input[name="amount"]').val();
var service_type = $j('input[name="item_name"]').val();
var order_to = $j('input[name="to"]').val();
var order_from = $j('input[name="from"]').val();
var order_message = $j('textarea#message').val();
if(hasError == false) {
var dataString = 'order_id='+ order_id + '&order_amount=' + order_amount + '&service_type=' + service_type + '&order_to=' + order_to + '&order_from=' + order_from + '&order_message=' + order_message;
$j.ajax({ type: "GET", cache: false, url: "/gc_process.php", data: dataString, success: function() { } });
} else {
return false;
}
});
Here's what my PHP script looks like:
<?php
// Make a MySQL Connection
include('dbconnect.php');
// Get data
$order_id = $_GET['order_id'];
$amount = $_GET['order_amount'];
$type = $_GET['service_type'];
$to = $_GET['order_to'];
$from = $_GET['order_from'];
$message = $_GET['order_message'];
// Insert a row of information into the table
mysql_query("REPLACE INTO gift_certificates (order_id, order_type, amount, order_to, order_from, order_message) VALUES('$order_id', '$type', '$amount', '$to', '$from', '$message')");
mysql_close();
?>
Any ideas?

Answer 1

You really should be using POST instead of GET, but regardless, I would check the following:
That jQuery is executing the ajax call after you click back and change the information, you should probably put either a console.log or an alert calls to see if javascript is failing
Add some echos in the PHP and some exits and go line by line and see how far it gets. Since you have it as a get, you can just load up another tab in your browser and change the information you need to.
if $j in your jQuery is the form you should be able to just do $j.serialize(), it's a handy function to get all the form data in one string

Answer 2

Mate,
Have you enclosed your jquery in
$j(function(){
});
To make sure it is only executed when the dom is ready?
Also, I'm assuming that you've manually gone and renamed jquery from "$" to "$j" to prevent namespace conflicts. If that isn't the case it should be $(function and not $j(function
Anyway apart from that, here are some tips for your code:
Step 1: rename all the "name" fields to be the name you want them to be in your "dataString" object. For example change input[name=from] to have the name "order_from"
Step 2:
Use this code.
$j(function(){
$j(".submit").click(function() {
var hasError = false;
if(hasError == false) {
var dataString = $j('form').serialize();
$j.ajax({ type: "GET", cache: false, url: "/gc_process.php?uu="+Math.random(), data: dataString, success: function() { } });
} else {
return false;
}
});
});
You'll notice i slapped a random variable "uu=random" on the url, this is generally a built in function to jquery, but to make sure it isn't caching the response you can force it using this method.
good luck. If that doesn't work, try the script without renaming jquery on a fresh page. See if that works, you might have some collisions between that and other scripts on the page

Answer 3

Turns out the problem is due to the fact that I am using iframes. I was able to fix the problem by making the page without iframes. Thanks for your help all!