Remove All Unnecessary Whitespaces from JSON String (in PHP)

Question

How do I remove ALL unnecessary whitespaces from a JSON String (in PHP)?

I assume that I need to use preg_replace with some clever regex in order to NOT touch the whitespaces that are part of the values.

A simple example would be:

Before: '{ "key": "value with whitespaces to maintain" }'

After: '{"key":"value with whitespaces to maintain"}'

Basically, I'm looking for a way to minify and pack the string as tight as possible without changing any data.

One way would be to parse JSON and then use your own encoding implementation, it's not very difficult. — Cthulhu, May 25 '14 at 20:38

score 22 · Accepted Answer · answered May 25 '14 at 20:46

22

Sorry to state the obvious:

$before = '{ "key": "value with whitespaces to maintain" }';
$after  = json_encode(json_decode($before));

And it actually matches perfectly your example, see $after:

{"key":"value with whitespaces to maintain"}

answered May 25 '14 at 20:46

hakre

What if `$before` is array or object? Nested function calling doesn't seem to work... – Lukas Feb 25 '23 at 12:40
@Lukas: Yes, given `$before` is not JSON Text but a PHP array or object value, there is no need to nest the function calls, as you only need to encode and not decode: `$after = json_encode($before);`. That's all, HTH. – hakre Mar 02 '23 at 13:12

bloodyKnuckles · Answer 2 · 2021-11-02T18:03:17.667

5

A PHP preg_ solution:

preg_replace(
    '/\s(?=([^"]*"[^"]*")*[^"]*$)/', ''
    , '{ "key": "value a with whitespaces to maintain" }'
);

edited Nov 02 '21 at 18:03

answered May 25 '14 at 21:34

bloodyKnuckles

1

very nice. I used it to cleanup qraphql queries before sending them to the API – MoVod Oct 10 '21 at 01:37

score 0 · Answer 3 · answered Jan 26 '17 at 11:18

PHP =>

Syntax : ltrim(string,charlist)

Example :

`$str = '{ "name" : " Test Subject" }';`
`$obj = json_decode($str);`
`$obj->name = ltrim($obj->name);`
`var_dump($obj);`

JS/jQuery =>

Syntax : jQuery.trim( str )

Example :

`var obj={ "name" : " Test Subject" };`
`console.log(obj);`
`obj["name"]=obj.name.trim(); // OR // obj.name.replace(/^\s+/,"");`
`console.log(obj);`

3 Answers3