How to log pretty printed json in PHP? - 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?

Answer 1

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.

Answer 2

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.

Answer 3

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.