Minify indented JSON string in .NET

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夕颜 2020-12-08 02:57

I have an indented JSON string e.g.

{
  \"a\": 1
}

However, I don\'t have the type of the instance to be serialized or deserialized.

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  • 2020-12-08 03:01

    That regex below works for me.

    string json = @"{ 
                      "window": {
                      "title": "Sample Konfabulator Widget",
                      "name": "main_window",
                      "width": 500,
                      "height": 500
                      } 
                    }"
    
    Regex.Replace(json, @"\s(?=([^""]*""[^""]*"")*[^""]*$)", string.Empty);
    

    I am using it on .NET Core app project. Regex is from that answer.

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  • 2020-12-08 03:17
    Regex.Replace(myJSON, "(\"(?:[^\"\\\\]|\\\\.)*\")|\\s+", "$1")
    

    should do it. It makes sure that strings that contain space characters are preserved, and all other space characters are discarded. All JSON keywords (false, true, null) have to be separated by commas or other punctuation so only white-space inside strings needs to be preserved.


    The first option (\"(?:[^\"\\\\]|\\\\.)*\") matches a double quoted string. The (...) mean that the output is captured and available in the replacement as $1. The [^\"\\\\] matches any character except a double quote or escape character \.

    Since matching occurs left-to-right, the second option, \s+ will not match space inside a string.

    So we match whole strings, and spaces outside strings. In the former case, $1 is the string token, and in the latter case $1 is the empty string because group 1 was not used.


    This works as intended because

    1. the only tokens in JSON that can contain spaces are double-quoted strings. There are no single-quoted strings or comments in JSON.
    2. the JSON grammar requires single-character punctuation between all multi-character tokens, so removing space will not merge tokens. In JavaScript, this could be problematic because space is required to break tokens; var x=0 is different from varx=0 and x - -(y) is different from x--(y).
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