AJAX POST to PHP mySQL query

Question

I am getting the proper log in the console but it isn't going to the database I cannot see anything wrong with my PHP, though. Can anyone see what might be the problem?

JQuery code:

 $( "#dialog-form" ).dialog({
      autoOpen: false,
      height: 300,
      width: 350,
      modal: true,
      buttons: {
        "Create an account": function() {

        var fname = $( "#guestfname" ).val();
        var lname = $( "#guestlname" ).val();
        var gender = $( "#guestgender" ).val();
        var address = $( "#guestaddress" ).val();
        var city = $( "#guestcity" ).val();
        var state = $( "#gueststate" ).val();
        var zip = $( "#guestzip" ).val();
        var phone = $( "#guestphone" ).val();
        var email = $( "#guestemail" ).val();
        var dob = $( "#guestdob" ).val();

        var dataString ={fname:fname, lname:lname, gender:gender, address:address,
                            city:city, state:state, zip:zip, phone:phone, email:email,
                            dob:dob};
        console.log(dataString);    
            $.ajax({        
                type: "POST",
                url: "classes/add_guest.php",
                data: dataString,
                cache: false,
                success: function(html)
                {
                    $('.guestinfo').html(html);
                }
            });
            $( this ).dialog( "close" );
        },
        Cancel: function() {
          $( this ).dialog( "close" );
        }
      },
      close: function() {
      }
    });

    $( "#create-user" )
      .button()
      .click(function() {
        $( "#dialog-form" ).dialog( "open" );
      });
  });

PHP mySQL query code:

<?php
//open connection
require_once('../config/db.php');
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

//get date
$fname = $_POST['fname'];
$lname =  $_POST['lname'];
$gender = $_POST['gender'];
$address = $_POST['address'];
$city = $_POST['city'];
$state = $_POST['state'];
$zip = $_POST['zip'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$dob = $_POST['dob'];


mysqli_query($con, 'INSERT INTO guests(fname, lname, gender, address, city, state, zip, phone, email, dob)
                    VALUES("'.$fname.'","'.$lname.'","'.$gender.'","'.$address.'","'.$city.'","'.$state.'",'.
                    $zip.','.$phone.',"'.$email.'",'.$dob.'');

mysqli_close($con);
?>

Machavity · Accepted Answer · 2013-10-30 00:14:49Z

1

First off, you're not escaping anything. Or using prepared statements. Hello, Bobby Tables (or SQL injection if you like).

Second, you're not wrapping your values in your statements in quotes. For instance, $phone isn't wrapped. I don't think MySQL would take too kindly to

INSERT INTO table(phone) VALUES(123-867-5309)

So we need to quote all your strings

INSERT INTO table(phone) VALUES("123-867-5309")

Third, What's going on with your AJAX? Did you use something like Firebug to make sure your AJAX was being sent properly? And have you ever heard of serialize? MUCH easier way to post your form data.

answered Oct 30, 2013 at 0:14

Machavity♦

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Comments

jacouh · Accepted Answer · 2013-10-30 00:16:47Z

1

There is a slight problem, string must be single quoted before passed to MySQL server:

mysqli_query($con, 'INSERT INTO guests(fname, lname, gender, address, city, state, zip, phone, email, dob)
                VALUES("'.$fname.'","'.$lname.'","'.$gender.'","'.$address.'","'.$city.'","'.$state.'",'.
                $zip.','.$phone.',"'.$email.'",'.$dob.'');

=> Good syntax here:

mysqli_query($con, "INSERT INTO guests(fname, lname, gender, address, city,
  state, zip, phone, email, dob)
VALUES('$fname', '$lname', '$gender', '$address', '$city',
  '$state', '$zip', '$phone', '$email', '$dob'");

edited Oct 30, 2013 at 0:16

answered Oct 30, 2013 at 0:10

jacouh

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Fallen · Accepted Answer · 2013-12-20 02:55:40Z

1

First of all, your query string doesn't need to be added together, PHP can pick out which are variables. So try this

mysqli_query($con, "INSERT INTO guests (fname, lname, gender, address, city, state, zip, phone, email, dob) VALUES ('$fname','$lname','$gender','$address','$city','$state','zip','$phone','$email','$dob')";

If that doesn't work add error_reporting(E_ALL); to the top of your page try changing all your $_POST's to $_REQUEST's. Then go to the page with the variables in the URL (GET Method). That should tell you the error. For example website.com/add_guest.php?lname=Dave

edited Dec 20, 2013 at 2:55

answered Oct 30, 2013 at 0:12

Fallen

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Burning Hippo · Accepted Answer · 2013-10-30 00:28:57Z

0

Thanks for the help all. We all missed it, though. mysqli_query($con, "INSERT INTO guests(fname, lname, gender, address, city, state, zip, phone, email, dob) VALUES('$fname', '$lname', '$gender', '$address', '$city', '$state', '$zip', '$phone', '$email', '$dob')");

Forgot that ) to close the VALUES.

answered Oct 30, 2013 at 0:28

Burning Hippo

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