Question

I have sent data to my jquery file and I tested the data string and everything is stored where it should be but when I post it to my php file all the variables are null. Can someone help me?

$(document).ready(function() {
  $(".button").click(function() {
  //$('.error').hide();
  var firstname = $("input#First_Name").val(); 
  var lastname = $("input#Last").val(); 
  var areacode = $("input#area_code").val(); 
  var phonenumber = $("input#Phone_Number").val(); 
  var emailaddress = $("input#emailaddress").val(); 
  var confirmemail = $("input#confirm_email").val(); 
  var password = $("input#Password_Input").val(); 
  var confirmpassword = $("input#ConfirmPassword").val(); 
  var streetaddress = $("input#StreetAddress_input").val(); 
  var streetaddress2 = $("input#StreetAddress2_input").val();
  var city = $("input#City_Input").val(); 
  var state = $("input#StateInput").val(); 
  var zipcode = $("input#ZipCode").val(); 
  var month = $("input#month_input").val(); 
  var day = $("input#day_input").val(); 
  var year = $("input#year_input").val(); 
  var services = $("input#services_input").val(); 
  var agreement = $("input#agreement").val(); 

  var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&areacode=' + areacode + '&phonenumber=' + phonenumber + '&emailaddress=' + emailaddress + '&confirmemail=' + confirmemail + '&password=' + password + '&streetaddress=' + streetaddress + '&streetaddress2=' + streetaddress2 + '&city=' + city + '&state=' + state + '&zipcode=' + zipcode + '&month=' + month + '&day=' + day + '&year=' + year + '&services=' + services + '&agreement=' + agreement; 
  alert(dataString); 
  $.ajax({
        type: "POST",
        url: "http://www.vectorcreditsolution.com/js/process.php",
        data: dataString,
            success: function() {
            alert("Yay it was sent"); 
    }
        });
        return false;
     });

and then my php file

<?php
    $FirstName = $_POST["firstname"];  
    $LastName = $_POST['lastname']; 
    $AreaCode = $_POST['areacode']; 
    $PhoneNumber = $_POST['phonenumber']; 
    $EmailAddress = $_POST['emailaddress']; 
    $Password = $_POST['password']; 
    $StreetAddress = $_POST['streetaddress']; 
    $StreetAddress2 = $_POST['streetaddress2']; 
    $City= $_POST['city']; 
    $State = $_POST['state']; 
    $ZipCode = $_POST['zipcode']; 
    $Month = $_POST['month']; 
    $Day = $_POST['day']; 
    $Year= $_POST['year'];
    $Service=$_POST['services']; 

    var_dump($_POST["firstname"]); 
    var_dump($_POST['firstname']); 
    var_dump($_POST[firstname]); 

Answer 1

You aren’t outputting the returned data from the AJAX and you aren’t using a tool to view it. Perhaps you are expecting the var dumps to appear on the page. They will not, they will be returned into the success function where you can use the data how you see fit.

jQuery’s AJAX method will pipe the output from your PHP file into the ‘ret’ variable inside the success function. If you wish to return an HTML success message, for example, you can then add that HTML to the page without refreshing it. Or any of a million things.

You should read up on AJAX.

Change This:

$.ajax({
type: "POST",
    url: "http://www.vectorcreditsolution.com/js/process.php",
    data: dataString,
    success: function() {
        alert("Yay it was sent"); 
    }
});

To This:

$.ajax({
type: "POST",
    url: "http://www.vectorcreditsolution.com/js/process.php",
    data: dataString,
    success: function(ret) {
        alert(ret); 
    }
});

And then click your button on the page with the jQuery.

Do not combine this with the other answers on this page. One is going to change your array keys on you (serialize).