How to log pretty printed json in PHP?

Question

I'm building an API in php. This API processes json messages from a third party API.

I want to log invalid pretty printed json messages.

So I did this:

error_log("test\n" . json_encode($json_string, JSON_PRETTY_PRINT));

However, when I look at my logs, the json string is not pretty printed:

$ tailf error.log
2015-07-13 10:20:03: (mod_fastcgi.c.2701) FastCGI-stderr: test
"{\"info\":{\"status\":200,\"msg\":\"OK\"},\"response\":{\"foo\":\"bar\"}"

I want to see something like:

$ tailf error.log
2015-07-13 10:20:03: (mod_fastcgi.c.2701) FastCGI-stderr: test
{
  "info": {
     "status": 200,
     "msg ": "OK"
  },
  "response": {
     "foo": "bar"
  }
}

How can I achive this result?

you need to add \n into your JSON... so, you need to parse your JSON after encoding to add \n after what you want.. so after "," and "}" and before } when "," not appear after. — Jean-philippe Emond, Jul 13 '15 at 14:38
http://stackoverflow.com/questions/352098/how-can-i-pretty-print-json?rq=1 — Marc B, Jul 13 '15 at 14:38
@Jean-philippeEmond Can you post an answer explaining your point please? — Stephan, Jul 13 '15 at 15:12
Use your own error handler instead of forcing the native one into a format which it is not supposed to. Btw, read the warning boxes on [php.net/error_log](http://php.net/manual/en/function.error-log.php) — Daniel W., Jul 13 '15 at 16:31

TRiG · Accepted Answer · 2015-07-13T16:36:33.820

error_log("test\n" . json_encode($json_string, JSON_PRETTY_PRINT));

json_encode() will actually not necessarily produce JSON: it will produce something that can be read by javascript. If you give it an array or an object, it will produce JSON; if you give it a string, it will produce a javascript string. And that’s what you’re doing, so that’s what you’re getting.

To be clear, $json_string is a string: (as far as PHP is concerned, it’s a string; if you passed that same string to javascript, it would be interpreted as an object). You pass that through json_encode() and all you’re going to get is another string (a string of doubly-encoded JSON).

JSON_PRETTY_PRINT is having no effect here, because you’re not producing JSON: you’re producing something that javascript too would see as a string.

Savvy?

So what you need to do is to (a) turn $json_string back into a PHP array, and then (b) reencode that as JSON, this time using the JSON_PRETTY_PRINT flag.

$log_array = json_decode($json_string, true);
$json_pretty_string = json_encode($log_array, JSON_PRETTY_PRINT);
error_log('test' . PHP_EOL . $json_pretty_string);

Rather than converting it back to a PHP array and then back to JSON, it would be better to add the JSON_PRETTY_PRINT flag to wherever you’re getting $json_string from in the first place, if possible.

Alternatively, just log $json_string directly (no need to encode it: it’s already a string, you can pass it to error_log() as it is), and worry about prettifying it only when you need to read your logs. This will make your logs considerably smaller.

What's the difference between JSON and "something that can be read by javascript"-from-a-function-called-json_encode ? lol :D The answer is inaccurate in my opinion. No offence. — Daniel W., Jul 13 '15 at 16:34
@DanFromGermany. "Like the reference JSON encoder, json_encode() will generate JSON that is a simple value (that is, neither an object nor an array) if given a string, integer, float or boolean as an input value. While most decoders will accept these values as valid JSON, some may not, as the specification is ambiguous on this point." [json_encode()](http://php.net/manual/en/function.json-encode.php) — TRiG, Jul 13 '15 at 16:47
My wording could perhaps be clarified. I'd say that the "simple value" output is not true JSON, but it seems that there's debate on that point. — TRiG, Jul 13 '15 at 16:48

score 1 · Answer 2 · answered Jul 13 '15 at 16:40

Common unix error logs are not supposed to contain human-readable json or other unescaped characters. Many syslog/logging implementations are limited by character width and automatically add encoding (like \") or remove new line characters, PHP's error_log is not binary safe either - the behaviour when encountering a unicode character is unpredictable tho (not sure).

You should not be using the native syslog/error log functions, instead, build your own logger, dedicated to json logging.

Personally I use MongoDB for logging json, because it's the kind of data MongoDB is supposed to work with.

Jean-philippe Emond · Answer 3 · 2015-07-13T16:31:36.957

0

You have two options in this case,

if you can use a str_replace like:

error_log("test\n" . str_replace('\"',"\n",json_encode($json_string, JSON_PRETTY_PRINT)));

or like @Karoly Horvath said:

You encode a string already encoded in a JSON. Your $json_string is already encoded. So you need to decode your first JSON and re-encode it with good parameters

error_log("test\n" . json_encode(json_decode($json_string), JSON_PRETTY_PRINT));

and give credit to @Karoly.

edited Jul 13 '15 at 16:31

answered Jul 13 '15 at 16:19

Jean-philippe Emond

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Why replace escaped quotation marks by newlines?? Btw, you should use the constant `PHP_EOL` instead of `\n` or `\r\n`, etc. – Daniel W. Jul 13 '15 at 16:27
Good point. its not the best answer. its just to explain a working code. the best answer is @Karoly Horvath. I did'nt see that is both json encoding. but I don't want to stole answer of people so, I explain mine like I said on the comment. ;-). For the PHP_EOL, I just forget this global. – Jean-philippe Emond Jul 13 '15 at 16:32

How to log pretty printed json in PHP?

3 Answers3