CODEWITHSUNDEEP
phpjquerymysqlajaxdatabase
Kyle Heier
Kyle Heier

Reputation: 15

AJAX form not inserting values into SQL DB

I am trying to incorporate this code to allow me to register a user after their details are given. All details are to be inserted into the database, and then load the new page, all done using AJAX.

To help keep things in context. All pages are loaded within the #main-content div within index.php. They are all loaded via the same function used that you will see in the click.js portion upon ajax success. Register.php is simply one of the pages that loads within this div.

Currently, the form loads properly, and upon submission, the new page loads as per the ajax function. Nothing however is inserted into the database.

Disclaimer: I have not set this up yet to trim for security purposes. This is a matter of function first prior to setting up to protect against SQL injections.

register.php

<script src="js/click.js"></script>

<form action="click.js" method="post">

   <label for="first_name">First Name:</label>
   <input type="text" id="first_name" name="first_name" /><br>

   <label for="last_name" >Last Name:</label>
   <input type="text" id="last_name" name="last_name" /><br>

   <label for="username">Username:</label>
   <input type="text" id="username" name="username" /><br>

   <label for="password">Password:</label>
   <input type="text" id="password" name="password" /><br>

   <label for="email">Email:</label>
   <input type="email" id="email" name="email" /><br>

   <button type="submit" id="reg-submit" name="submit">Submit</button>

</form>

click.js

$(document).ready(function(){

  $('#reg-submit').click(function() {
     var firstName = $('#first_name').val();
     var lastName = $('#last_name').val();
     var userName = $('#username').val();
     var password = $('#password').val();
     var email = $('#email').val();
     var dataString = 'name1=' + firstName + '&lastname1=' + lastName + '&user1=' + userName + '&password1=' + password + '&email1=' + email;

     if (firstName == "" || lastName == "" || userName == "" || password == "" || email == "") {
         alert('missing some information');
     }  else {
         $.ajax({
           type: "POST",
           url: "usersubmit.php",
           data: dataString,
           cache: false,
           success: function(){
             $('#main-content').load('php/next-page.php').hide().fadeIn('slow');                
          }
        });
      };
   return false;
   });
});

the DB connection takes place within the users_db.php.

usersubmit.php

<?php 

   include 'users_db.php';

   $first1=$_POST['name1'];
   $last1=$_POST['lastname1'];
   $username1=$_POST['user1'];
   $pass1=$_POST['password1'];
   $email01=$_POST['email1']; 

   $userinfo = $conn->prepare("INSERT INTO registered_users (FirstName, LastName, Username, Password, Email) VALUES ('$first1', '$last1', '$username1'', '$pass1', '$email01')");

   $userinfo->execute();


   $conn = null;
?>

Much appreciated!

If you see any other problems I may have here outside of the form simply not submitting, feel free to point them out.

Upvotes: 0

Views: 70

Answers (1)

Marko Mackic
Marko Mackic

Reputation: 2333

The answer is that is not how you prepare statements :) 

<?php 

   include 'users_db.php';

   $first1=$_POST['name1'];
   $last1=$_POST['lastname1'];
   $username1=$_POST['user1'];
   $pass1=$_POST['password1'];
   $email01=$_POST['email1']; 

   $userinfo = $conn->prepare("INSERT INTO registered_users (FirstName, LastName, Username, Password, Email) VALUES (?, ?, ?, ?, ?)");
   $userinfo->bind_param("sssss",$first1,$last1,$username1,$pass1,$email01);

   $userinfo->execute();

    // you shoud close the prep statement object
   $userinfo->close();
   //this is the way to kill the conn
   $conn->close();
?>

This is assuming your connection to database works :)

Upvotes: 2

Related Questions