Multiple Flags for json_encode()

Question

How do I use multiple flags for the php json_encode()-function?

json_encode($array, JSON_PRETTY_PRINT, JSON_UNESCAPED_UNICODE);

This doesn't work - as just the first flag will be done an the second will be ignored.

possible duplicate of [What are PHP flags in function arguments?](http://stackoverflow.com/q/9635301) (background info on how it works). — mario, Aug 31 '15 at 12:26
I'm voting to close this question as off-topic because the question is clearly answered through examples in the [official documentation](http://php.net/manual/en/function.json-encode.php) — Jocelyn, Aug 31 '15 at 13:58

score 78 · Accepted Answer · answered Aug 31 '15 at 12:24

78

json_encode($array, JSON_PRETTY_PRINT|JSON_UNESCAPED_UNICODE);

This will add the binary values of JSON_PRETTY_PRINT and JSON_UNESCAPED_UNICODE with the binary OR operator.

answered Aug 31 '15 at 12:24

Byte Welder

score 13 · Answer 2 · edited May 23 '17 at 12:25

13

Those flags are bitmasks. I wrote about it once a long time ago here on SO.

So, basically, to use more than one option, you need to or them together

json_encode($array, JSON_PRETTY_PRINT | JSON_UNESCAPED_UNICODE);

edited May 23 '17 at 12:25

Community

answered Aug 31 '15 at 12:25

Peter Bailey

2 Answers2