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POAJAX form wont post
 

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    <p>Hi everyone I am trying to submit a form without the page reloading, a seemingly straightforward thing with AJAX, but I can't seem to get any changes on my database. I have an alert set in my javascript to run if the AJAX was successful, and it seems to trigger every time, so I think the variables are, indeed, getting sent out of the AJAX, but when I get to join.php, I don't think it remembers, and therefore doesn't process correctly.</p> <p>First of all, here is the relevant script.js:</p> <pre><code>$(document).ready(function () { //----SUBMIT---// $(".submit").click(function () { // validate and process form here $('.error').hide(); var studentEmail = $("input#studentEmail").val(); if (studentEmail == "") { $("label#studentEmail_error").show(); $("input#studentEmail").focus(); return false; } var studentPassword = $("#studentPassword").val(); var parentEmail = $("#parentEmail").val(); var parentPassword = $("#parentPassword").val(); var studentFirstName = $("#studentFirstName").val(); var studentLastName = $("#studentLastName").val(); var studentPhone = $("#studentPhone").val(); var parentFirstName = $("#parentFirstName").val(); var parentLastName = $("#parentLastName").val(); var parentPhone = $("#parentPhone").val(); var dataString = 'studentEmail=' + studentEmail + '&amp;studentPassword=' + studentPassword + '&amp;parentEmail=' + parentEmail + '&amp;parentPassword=' + parentPassword + '&amp;studentFirstName=' + studentFirstName + '&amp;studentLastName=' + studentLastName + '&amp;studentPhone=' + studentPhone + '&amp;parentFirstName=' + parentFirstName + '&amp;parentLastName=' + parentLastName + '&amp;parentPhone=' + parentPhone; alert(dataString); $.ajax({ type: "POST", url: "join.php", data: dataString, success: function () { alert("success"); } }); return false; }); }); </code></pre> <p>Then join.php</p> <pre><code>if($_POST) { $studentEmail=$_POST['studentEmail']; $studentPassword=$_POST['studentPassword']; $parentEmail=$_POST['parentEmail']; $parentPassword=$_POST['parentPassword']; $studentFirstName=$_POST['studentFirstName']; $studentLastName=$_POST['studentLastName']; $studentPhone=$_POST['studentPhone']; $parentFirstName=$_POST['parentFirstName']; $parentLastName=$_POST['parentLastName']; $parentPhone=$_POST['parentPhone']; //create database connection $connection = mysql_connect("localhost","XXXX","XXXX"); //in case database connection fails if(!$connection) { die("Database connection failed: ".mysql_error()); } else{ //select database to use $db_select = mysql_select_db("XXXX",$connection); //in case database selection fails if (!$db_select) { die("Database selection failed: " . mysql_error()); } else { //make sql query $sql = "INSERT INTO clients (`studentEmail`, `studentPassword`, `parentEmail`, `parentPassword`, `studentFirstName`, `studentLastName`, `studentPhone`, `parentFirstName`, `parentLastName`, `parentPhone`) VALUES ('".$studentEmail."', '".$studentPassword"', '".$parentEmail."', '".$parentPassword."', '".$studentFirstName."', '".$studentLastName."', '".$studentPhone."', '".$parentFirstName."', '".$parentLastName."', '".$parentPhone."')"; //set results to variables $result = mysql_query($sql); //in case query fails if (!$result) { die("Database query failed: " . mysql_error()); } } } } else { echo "FAIL"; } </code></pre>
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    1. COI know this doesn't answer your question, but a simple way to get form data in jQuery is to use `$('yourForm').serialize()`. Thought I'd pass the tip along.
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