Alfraganus Posted August 19, 2015 Share Posted August 19, 2015 i dont know why but, ajax is not sending request to database, I supposed to register email with the help of database, let me show codes, when I am clicking submit button, nothing is happening. HTML---- <html> <head> <script src="http://code.jquery.com/jquery-2.1.4.js"></script> <script> $(document).ready(function(){ $("#sub").click(function(){ var name = $("#name").val(); var email = $("#email").val(); var pass = $("#pass").val(); $.POST('test.php',{ name:name,email:email,pass:pass },function(data){ $("#result").html(data); }); }); }); </script> </head> <body> <div id="box"> <h2>new user register</h2> <input type="text" name="name" id="name" placeholder="enter your name"/><br> <input type="text" name="email" id="email" placeholder="enter your email"/><br> <input type="password" name="pass" id="pass"/><br> <input type="submit" name="sub" value="submit" id="sub" > <div id="result"> </div> </div> </body> </html> ----test.php............ <?php $con=mysqli_connect("localhost","root","","test"); $name=$_POST ['name']; $email=$_POST ['email']; $pass=$_POST ['pass']; $sel="select * from users where email='$email'"; $run=mysqli_query ($con, $sel); $check_email=mysqli_num_rows ($run); if ($check_email==1) { echo "<h2>This is email is already registered, try another</h2>"; exit(); } else { $insert="insert into users (name,pass,email) values ('$name','$pass',$email')"; $run_insert=mysqli_query($con,$insert); if($run_insert) { echo "<h2>registration has been succesfull G'ulomjon aka </h2>"; } } ?> Link to comment Share on other sites More sharing options...
Ingolme Posted August 19, 2015 Share Posted August 19, 2015 Press F12 to open the Javascript console, then click the button and see what shows up there. Your code will fail if somebody types ' OR 1 OR ' as their e-mail address. You should research SQL injection. Link to comment Share on other sites More sharing options...
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