Error using json_encode in php

Question

I'm having some trouble when I try to use php's JSON encode function

what I'm doing:

echo json_encode($ppar, JSON_PRETTY_PRINT);

now the variable $ppar is an associative array with a whole bunch of data in it. I can get it to work with out the added parameter, but not with it, and when I looked up on php.net it does say there that the 2nd parameter is valid so I don't get what I'm doing wrong.

this works (except for the fact everything is squished together):

echo json_encode($ppar);

but this:

echo json_encode($ppar, JSON_PRETTY_PRINT);

gives me this error:

Warning: json_encode() expects exactly 1 parameter, 2 given in /home/www/mysite/pp.php on line 10

and the output is null. I've been going to this for reference: http://php.net/manual/en/function.json-encode.php

Rene Pot · Accepted Answer · 2013-07-04 19:10:46Z

5

You must be running a PHP version prior to 5.3. Since the doc states that is when it was introduced.

Also, JSON_PRETTY_PRINT was added in 5.4. So even though options will work on 5.3, PRETTY_PRINT won't

answered Jul 4, 2013 at 19:10

Rene Pot

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1 Comment

Kit Ramos Over a year ago

I just noticed it was using 5.2 I was able to get my host to upgrade to 5.3.6 and I no longer get the error message but still a null output. so I'm going to go with the answer that it was introduced in 5.4

AMADANON Inc. · Accepted Answer · 2013-07-04 19:10:50Z

3

The "options" parameter (which you set to JSON_PRETTY_PRINT) was added in PHP 5.3.0 - you are probably using a version of PHP older than that.

answered Jul 4, 2013 at 19:10

AMADANON Inc.

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