JSON not working in php script

Question

I'm not getting the json_encode to work in my php file. Like for instance, I tried this example I got from php.net

<?php
$arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);

echo json_encode($arr);
?>

But nothing works. If i remove the echo statement, then my php works, which means php isn't recognizing the json_encode code.

I'm using PHP 5.4.16. To sum it up, I'm using xampp 1.8.2.

Help please?

What exactly is happening when you say "nothing works"? Are you getting an error, or no output? — StephenTG, Aug 12 '13 at 19:45
Where is this PHP block in your page ? Is that the whole page ? — Denys Séguret, Aug 12 '13 at 19:46
Try using `json_last_error()` to see what error you're getting. http://www.php.net/manual/en/function.json-last-error.php — Tricky12, Aug 12 '13 at 19:46
Works fine for me: http://codepad.org/sq1LmgZb. If you have a problem, you should describe it properly. Especially you have to provide code that *reproduces* the problem. — Felix Kling, Aug 12 '13 at 19:47
@KAy what are you seeing when you remove the json_encode call? — Orangepill, Aug 12 '13 at 19:47
Add `ini_set('display_errors', 1); error_reporting(-1);` to the top of your file. — gen_Eric, Aug 12 '13 at 19:48
@dystroy its at the end. The rest of the code works normally, but just can't use the json code at all. — Kay, Aug 12 '13 at 19:48
What does it mean, "at the end" ? How is it supposed to be rendered ? — Denys Séguret, Aug 12 '13 at 19:50
You should check PHP/Apache log files. Maybe something is wrong with your configuration. Like missing some essential element. You would have the answer in the logs. — Thibault, Aug 12 '13 at 19:50
@Thibault thanks for the suggestion. This is what it says in the log: Can't use function return value in write context — Kay, Aug 12 '13 at 19:54
Is your script used like this as standalone ? Is it complete ? Is it called by another script like eval or something ? — Thibault, Aug 12 '13 at 20:05
standalone. I just used that as a test to see if JSON is working. After that statement i have something like echo" alert('works');"; to make sure it works but the message never alerts unless I remove the json_encode statement — Kay, Aug 12 '13 at 20:07
@Kay: `Can't use function return value in write context`. That means you have `isset(func())` or `empty(func())`; you can't do that. They only work on variables. — gen_Eric, Aug 13 '13 at 14:32

score 1 · Answer 1 · answered Aug 12 '13 at 19:49

1

It sounds like there could be a fatal error somewhere else in your script. Make sure display_errors is set to On in php.ini.

answered Aug 12 '13 at 19:49

John Himmelman

21,504
22
65
80

score 0 · Answer 2 · edited May 23 '17 at 10:25

This question can not be answered without more debug beeing provided.

As in the comments suggested, use these lines before everything to (try to) enable error reporting:

<?php
header('Content-Type: text/plain');
error_reporting(E_ALL);
ini_set('display_errors', true);
ini_set('display_startup_errors', true);
// Json
echo json_last_error();

Use phpinfo(); to obtain information about PHP Version and JSON support.

This snippet will tell you if the json function is available:

<?php
if (!function_exists('json_encode')) {
    echo 'PHP not compiled with json support', PHP_EOL;
}

Last but not least, check errors logs

/var/log/httpd*
/var/log/nginx*
/var/log/php-fpm*
/var/log/apache*
.... (linux)

From Where does PHP's error log reside in XAMPP?

\xampp\apache\logs\error.log

or

\xampp\php\logs\php_error_log

Good advice except for the header() function. A) script might have already echoed something to output buffer causing more errors/confusion, B) many configurations use HTML formatted error messages, B) no real point in end — JSON, Aug 13 '13 at 22:07

JSON not working in php script

2 Answers2