Return json on exit php

Question

I have been using $.getJSON in order to verify if a script has done its job properly, using exit("{'status':true}") (or false) but now I need the script to return a value (or an array). I see that I can't use exit('{ "status": $myarray}');. What can I use instead? I am new to php- is it possible to do something like return '{ "status": $myarray}'; or something alike? Thanks in advance

score 5 · Answer 1 · answered May 23 '13 at 12:03

5

You can use the following function:

json_encode($data);

See: http://php.net/manual/en/function.json-encode.php

answered May 23 '13 at 12:03

Rijndael

3,683
2
24
26

And exit($data)? How will the ajax recieve the response? – Mihai Bujanca May 23 '13 at 12:04

score 4 · Answer 2 · edited May 23 '17 at 12:22

4

In your php, use json_encode as such:

<?php
    header('Content-Type: application/json');
    echo json_encode($myarray);
    // or, `exit(json_encode($myarray))` if global scope (see *REF)
?>

and in your jQuery use getJSON normally:

$.getJSON("test.php",function(result){
    ...
});

(*) REF: PHP - exit or return which is better?

edited May 23 '17 at 12:22

Community

1
1

answered May 23 '13 at 12:23

mg007

2,888
24
29

`header('Content-Type: application/json')` did the trick, thank you! – Manas Mar 01 '23 at 08:04

score 2 · Answer 3 · answered May 23 '13 at 12:27

2

If you want return data just like as array or single data in json then Try this code

In php file write like this

$value = 'welcome';
echo json_encode($value);exit;

or

$value = array("Saab","Volvo","BMW","Toyota");
print_r(json_encode($value));exit;

answered May 23 '13 at 12:27

Ashish Jain

760
1
8
23

user2362083 · Answer 4 · 2013-05-23T12:09:00.893

1

   `$jsonString = 'your json string'`
    $jsonArray = json_encode($jsonString);
    return $jsonArray

edited May 23 '13 at 12:09

answered May 23 '13 at 12:04

user2362083

33
7

1

I think the question is about encoding not decoding. – Beetroot-Beetroot May 23 '13 at 12:06

score 0 · Answer 5 · answered May 23 '13 at 12:33

I will post the way I did it, seems quite simple to me now. Thank you for your answers

$json_array=array('status'=>$row);
exit(json_encode($json_array));

AJAX call:

   $.getJSON( "savefunctions/getVideos.php", function(response) {
                         if( response.status ) alert(response.status);
                         });

Return json on exit php

5 Answers5