Using JSON, passing variable from php to javascript

Question

I wonder why is it not working?

check_login.php

<?php
session_start();
$data = array("username" => "true");
echo json_encode($data);
?>

my js file var linkName;

$.ajax({
 type: "POST",
 url: "check_login.php",
 dataType: "json",
 success: function(json){
 if(json.username != "true")
 {
 //do something
 }
 }
});

I am trying to get the username after checking whether or not the user has signed in yet in the php file, something like passing a session variable. But currently passing a string seems to already have a problem. Any know what I did wrong here?

Still not working the code above. Anyone want to help me out here?

Answer 1

You have an error in your PHP code. The square bracket ] is not needed in line 3. And, of course you should use echo json_encode($data);

• markz: You should check the response of the server with a tool like Firebug for Firefox or Developer Tools in Google Chrome. That way you can see if the server outputs anything or not.

Answer 2

 `if(json.data.username != "true")` is wrong you do not have data object
 `if(json.username != "true")` try this

Answer 3

You should echo the json data so that the page you get using ajax contains the string data.

<?php
session_start();
$data = array("username" => "true");
echo(json_encode($data));
?>
$.ajax({
 type: "POST",
 url: "check_login.php",
 dataType: "json",
 success: function(json){
 console.log(json)
 }
});

Also some times you might need to set the content type of the header in php to json.

• Pablo Jomer: Open the console in chrome or mozilla to view the console.log this is usualy done by clicking right any where in the webbrowser and then selecting inspect element. after that you will find a tab called console where logs appear.
• Pablo Jomer: Have you tried surfing in to check_login.php and looking at that page what dose that show???

Answer 4

try to send the header

header("Content-Type: application/json", true);
print json_encode($data);

Answer 5

 return json_encode($data);

I think you have to return the data rather than just json encoding it!

• Niklas Modess: return will not echo the output