Checking if username exists in Database jQuery

Question

I want to check if the username is already taken, here is my script, which outputs "undefined". Can anyone help me, please? :)

This is in my jQuery - $("#registerusername").val() is the value of an input.

$.post('checkregister.php',{username: $("#registerusername").val()}, function(data){
    window.alert(data.exists);
    if(data.exists){
        window.alert("Name already found");
    }else{
        window.alert("Name NOT found");
    }
}, 'JSON');

This is in my checkregister.php

header('content-type: text/json');
if(!isset($_POST['username']))
exit;
$db = new PDO('mysql:host=localhost;dbname=testdatabase','root','pw000');
$query = $db->prepare("SELECT * FROM users WHERE username = '" . $_POST['username'] . "'");
$query->execute();
echo json_encode(array('exists' => $query->rowCount() > 0));

Just thought you should know that this is prone to SQL injection. — Cᴏʀʏ
– Cᴏʀʏ, Commented May 8, 2013 at 18:22
See here on how to properly use prepared statements to avoid SQL injection attacks. — Cᴏʀʏ
– Cᴏʀʏ, Commented May 8, 2013 at 18:34

DMortensen · Accepted Answer · 2013-05-08 18:27:04Z

2

First, You might want to strengthen your php against sql injection by 'sanitizing' the input.

Next why return JSON from the php? It would be much simpler to just return either true or false.

$db = new PDO('mysql:host=localhost;dbname=testdatabase','root','pw000');
$query = $db->prepare("SELECT * FROM users WHERE username = '" . $_POST['username'] . "'");
$query->execute();
if( $query->rowCount() > 0 ){
    echo 'true';
}
else{
    echo 'false';
}

Then in your javascript:

$.post('checkregister.php',{username: $("#registerusername").val()}, function(data){
window.alert(data);
if(data == 'true'){
    window.alert("Name already found");
}else{
    window.alert("Name NOT found");
}
});

edit--- You could also just return a boolean variable from php rather than a string, but either will work

answered May 8, 2013 at 18:27

DMortensen

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3 Comments

Cᴏʀʏ Over a year ago

Better than sanitizing would be using prepared statements correctly.

Daan Over a year ago

Thank you so much, it's now working, and I learned something new :)

DMortensen Over a year ago

@Cory very true, but an easy implementation of mysql_real_escape_string will do wonders stackoverflow.com/questions/129677/…

sravis · Accepted Answer · 2013-05-08 18:43:30Z

Simple Example..

Jquery

var username  = $.trim($('#username').val());

if(username != '') {
$.ajax({
url : 'localhost/phpScript.php',
data : {username : username},
dataType : 'JSON',
type : 'POST',
cache : false,

success : function(result) {
if(result == '1') { alert('Username Found.'); }
else if(result == '0') { alert('Username Not Found!'); }
},

error : function(err) {
console.log(err);
}
});
}

PHP/MySQL (Make sure that you escape value for user input, Before giving it to MySql)

if(isset($_POST['username'])) {    
$username = $_POST['username'];
$sql = "SELECT username FROM users WHERE username = '".$username."' LIMIT 1";
$query = mysql_query($sql);
if(mysql_num_rows($query) == '1') { 
echo '1'; 
} else { 
echo '0'; 
}
}

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