How to show jsonp response sent by php on a page

Question

i have a script for jsonp as follows

<script>
$(document).ready(function(){

    $("#LoginForm").submit(function(){

    var data = $(this).serialize();
    //alert(data);
    $.ajax({
            type:"POST",
            dataType:"jsonp",
            url:"https://akshay.tk/api/login.php",
            data:data,

            success:function(data)
            {       

                /// WHAT TO WRITE HER TO GET PHP RESPONSE
                /// DIV WHERE DATA TO BE SHOWN, ID IS RESULT    
            }

            });
    return false;
    });

});
</script>

and my php code is

$username = $_REQUEST['username'];
$password = $_REQUEST['password'];              

// login check for user
$result = mysqli_query($con,"SELECT * FROM `admin_login` WHERE `username`='$username' AND `password`='$password'"); 

if(mysqli_num_rows($result) > 0) 
{
    $user = mysqli_fetch_assoc($result);
    // user found           
    $response["success"] = 1;
    $response["uid"] = $user["id"];
    $response["username"] = $user["username"];
    $response["role"] = $user["role"];

    echo json_encode($response);
}

Everything is going good and when i used developers tool then i came to know it is giving proper response as well. Response by php :

{"success":1,"uid":"1","username":"admin","role":"admin"}

What code should i write in jQuery SUCCESS function so that i can get php response?

http://stackoverflow.com/questions/6809053/simple-jquery-php-and-jsonp-example — Mujnoi Gyula Tamas, Nov 14 '15 at 08:57

score 1 · Accepted Answer · answered Nov 14 '15 at 09:05

1

If you want to ouput the result thrown by php for your ajax result then

Create a div like this

<div id='yourDiv'></div>

Then inside the success event

 success:function(data)
{       
('#yourDiv').html(data);
}

Note :

If you prefer class then

<div class='yourDiv'></div>

replaced by

('.yourDiv').html(data);

Additional Data :

It's better to check the data in your success event like this

As you getting like this as response

{"success":1,"uid":"1","username":"admin","role":"admin"}

success:function(data)
    {        
     if(data.success==1)
    {
    ('#yourDiv').html('Welcome'+data.username);  //You can append or do anything that you wish
    }
    else
    {
    ('#yourDiv').html.('Fail');
    }
    }

answered Nov 14 '15 at 09:05

Sulthan Allaudeen

11,330
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yes tried the same but not working...... i am tring this since morning. But error. its showing on browser developers tool but not on webpage – Akshay Nov 14 '15 at 09:22
If you have `id` on same name it won't work. you shall try class. – Sulthan Allaudeen Nov 14 '15 at 09:27
Also in the success event can you alert something ? – Sulthan Allaudeen Nov 14 '15 at 09:28
no i am not able to alert anything in success event – Akshay Nov 14 '15 at 09:39
Atleast inside `$("#LoginForm").submit(function(){` ? – Sulthan Allaudeen Nov 14 '15 at 09:41
yup in submit function it is prompting data but on success event its not prompting anything. i think its not going inside that function. i tried alert for ready and submit function. its working but for success its not working – Akshay Nov 14 '15 at 09:42
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/95116/discussion-between-akshay-and-sulthan-allaudeen). – Akshay Nov 14 '15 at 09:43

score 0 · Answer 2 · answered Nov 15 '15 at 05:00

Lots of RND i have got the answer jQuery code :

$("#submit").click(function(){

var myData = $("#LoginForm").serialize();

    $.ajax({
          method: "POST",
          dataType:"jsonp",
          url:"https://akshay.tk/api/login.php?callback=?",
          data: myData,

            success:function(msg)
            {
                alert(msg);                 
            }
        });

        return false;           
});

And Slight change in PHP code.

if(mysqli_num_rows($result) > 0) 
            {
                $user = mysqli_fetch_assoc($result);
                // user found           
                $response["success"] = 1;
                $response["uid"] = $user["id"];
                $response["username"] = $user["username"];
                $response["role"] = $user["role"];

                echo $_GET['callback'] . '(' . json_encode($response) . ')';
            }

And i t worked. I think callback function gets internally call as a reference or something so we can work on cross domain

Thank you

How to show jsonp response sent by php on a page

2 Answers2