How to check Email already exist in db using ajax?

Question

I am after a code for getting ajax response for checking whether an email address is already entered in the database. Here is my code snippet.

Java Script:

<script type="text/javascript">
pic1 = new Image(16, 16);
pic1.src="img/loader.gif";

$(document).ready(function(){

$("#email").change(function() {

var usr = $("#email").val();

if(usr.length >= 4)
{
$("#status").html('<img style="width:17px;" src="img/loader.gif" align="absmiddle">&nbsp;Checking availability...');

    $.ajax({ 
    type: "POST", 
    url: "check.php", 
    data: "email="+ usr, 
    success: function(msg){ 

   $("#status").ajaxComplete(function(event, request, settings){

    if(msg == 'OK')

    {
     alert ("error");

        $("#email").removeClass('object_error'); // if necessary
        $("#email").addClass("object_ok");
        $(this).html('&nbsp;<img src="tick.gif" align="absmiddle">');
    } 
    else 
    { 
        $("#email").removeClass('object_ok'); // if necessary
        $("#email").addClass("object_error");
        $(this).html(msg);
    } 

   });

 }

  });

}
else
    {
    $("#status").html('<font color="red">' +
'Enter Valid Email</font>');
    $("#email").removeClass('object_ok'); // if necessary
    $("#email").addClass("object_error");
    }

});

});
</script>

My Check.php file is as follows

<?php
// This is a sample code in case you wish to check the username from a mysql db table

if(isSet($_POST['email'])) {
$email = $_POST['email'];

$dbHost = 'localhost'; // usually localhost
$dbUsername = 'root';
$dbPassword = '';
$dbDatabase = 'jaquar_cdb';

$db = mysql_connect($dbHost, $dbUsername, $dbPassword)
 or die ("Unable to connect to Database Server.");
mysql_select_db ($dbDatabase, $db)
 or die ("Could not select database.");

$sql_check = mysql_query("select email from jaquar_contest where email='".$email."'")
 or die(mysql_error());

if(mysql_num_rows($sql_check)) {
    echo '<font color="red">The email <strong>'.$email.'</strong>'.
' is already in use.</font>';
} else {
    echo 'OK';
}
}
?>

Finally the HTML code

<div class="sf_columns column_3">                     
 <input id="email" type="text" placeholder="Email" name="email" data-required="true" data-email="true">
 <div style="left:0px;  padding-top: 10px;" id="status"></div>
</div>

The problem is that, I am not getting any response.. What is my mistake? Kindly help me..

score 4 · Answer 1 · edited May 23 '17 at 12:14

The problem why not getting any response is here with AjaxComplete function

$("#status").ajaxComplete(function(event, request, settings)

I'm not sure why you are using it but you can do without it and infact even it's not necessary with your approach

without Ajaxcomplete function made some changes in code.

pic1 = new Image(16, 16);
pic1.src="img/loader.gif";
$(document).ready(function(){
    $("#email").change(function() {
    var usr = $("#email").val();
    if(usr.length >= 4){
        $("#status").html('<img style="width:17px;" src="img/loader.gif" align="absmiddle">&nbsp;Checking availability...');
        $.ajax({ 
        type: "POST", 
        url: "check.php", 
        data: "email="+ usr,
        dataType: 'text',
            success: function(msg){
                if(msg == 'OK'){
                 alert ("success");
                    $("#email").removeClass('object_error'); // if necessary
                    $("#email").addClass("object_ok");
                    $("#status").html('&nbsp;<img src="tick.gif" align="absmiddle">');
                } else {
                    alert ("error");
                   $("#email").removeClass('object_ok'); // if necessary
                   $("#email").addClass("object_error");
                   $("#status").html(msg);
               }
            }

        });
    } else {
        $("#status").html('<font color="red">' + 'Enter Valid Email</font>');
        $("#email").removeClass('object_ok'); // if necessary
        $("#email").addClass("object_error");
    }
    });
});

SideNote: mysql_* functions is deprecated, so should stop using it and start using MySQLi

<?php
// This is a sample code in case you wish to check the username from a mysql db table
if(isset($_POST['email'])) { //change isSet to isset (it will not make any difference)
    $email = mysql_real_escape_string($_POST['email']); //escape the string

    $dbHost = 'localhost'; // usually localhost
    $dbUsername = 'root';
    $dbPassword = '';
    $dbDatabase = 'jaquar_cdb';
    $db = mysql_connect($dbHost, $dbUsername, $dbPassword)
        or die ("Unable to connect to Database Server.");
    mysql_select_db ($dbDatabase, $db)
        or die ("Could not select database.");

    $sql_check = mysql_query("SELECT email FROM jaquar_contest WHERE email='$email'") or die(mysql_error());
        if(mysql_num_rows($sql_check) > 0) { //check rows greater then zero (although it will also not make any difference)
            echo '<font color="red">The email <strong>'.$email.'</strong>'. ' is already in use.</font>';
        } else {
            echo 'OK';
        }
}
?>

you are welcome, don't forget to mark the answer correct :) – Shehary Oct 10 '15 at 03:49 — Shehary, Oct 10 '15 at 03:49

score 0 · Answer 2 · answered Oct 03 '15 at 16:02

Change your first line of php code to :

if(isset($_POST['email']))

And so your updated Check.php file:

<?php
// This is a sample code in case you wish to check the username from a mysql db table

if(isset($_POST['email'])) {
$email = $_POST['email'];

$dbHost = 'localhost'; // usually localhost
$dbUsername = 'root';
$dbPassword = '';
$dbDatabase = 'jaquar_cdb';

$db = mysql_connect($dbHost, $dbUsername, $dbPassword)
 or die ("Unable to connect to Database Server.");
mysql_select_db ($dbDatabase, $db)
 or die ("Could not select database.");

$sql_check = mysql_query("select email from jaquar_contest where email='".$email."'")
 or die(mysql_error());

if(mysql_num_rows($sql_check)) {
    echo '<font color="red">The email <strong>'.$email.'</strong>'.
' is already in use.</font>';
} else {
    echo 'OK';
}
}
?>

How to check Email already exist in db using ajax?

2 Answers2