CODEWITHSUNDEEP
phpjqueryjson
dr_rk
dr_rk

Reputation: 4565

$.getJSON fails on valid JSON?

I am pulling out my hair on JQuery/JSON issue where the response of $.getJSON function is not making sense.

So here is my Javascript code excerpt: 

var jqxhr = $.getJSON("myPHPFunction.php", function() {
      alert("success");
    })
    .success(function() { alert("second success"); })
    .error(function() { alert("error"); })
    .complete(function() { alert("complete"); });

The JSON returned by the PHP function is this: 

{"sid":"12", "name":"somename", "email":"someemail"}

Full response:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head><title>Untitled Document</title><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body> {"sid":"12", "name":"somename", "email":"someemail"}
</body>
</html>

When I run the script, I get a dialog box saying error.

Several online validator have reported the JSON to be syntax-correct. Then why does JQuery fail? Note both the php and the script are in the same folder (on server) and same domain.

Upvotes: 0

Views: 179

Answers (2)

turiyag
turiyag

Reputation: 2887

From the comments, you can't encapsulate the JSON in HTML markup. You'll need to ensure that the data returned ("View Source" is a good tool for verifying this) is pure JSON.

Upvotes: 3

rekire
rekire

Reputation: 47955

Your response must not contain any HTML!

Propberbly you also need to set the correct content type:

header("Content-Type: application/json");
echo('{"sid":"12", "name":"somename", "email":"someemail"}');
exit();

The code above should work.

Please note that you must not output any other code before the header line.

Upvotes: 1

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