正则表达式/JavaScript：拆分字符串以按每行最大字符数分隔行，并向后查找 n 个字符以寻找可能的空格？

Question

This is similar question to How to split a string at every n characters or to nearest previous space , however, on the contrary to what I was expecting based on the title, that solution does not work if there is just one long word without any whitespace.

So I need a Regex which splits a string to separate lines (multiple times if needed) by max characters per line , and looking backwards n characters for a possible whitespace (break there if found, otherwise at max length)?

Edit 1: For example, with max line length 30 characters with 15 characters backwards whitespace lookup:

Loremipsumissimplydummytextofthe printing and typesetting industry.

That sentence's first word has a length of 32 characters. So the output should be:

Loremipsumissimplydummytextoft  # Line has length of 30 char
he printing and typesetting     # Cut before the word at otherwise 30 char
industry.

So the first word should be force-cut after 30th character, as there was no whitespace.

The remaining string has a length of 28 (or 29 with the dash) before word 'industry', so at the place of 30th character there's a word, so the solution looks up for the previous whitespace within 15 characters range. That line is broken before 'industry' word.

Edit 2: Second example of text:

Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry. Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry.

Should output:

Loremipsumissimplydummytextoft
he printing and typesetting
industry. Loremipsumis simply
dummytext ofthe printing and
typesetting industry.
Loremipsumissimplydummytextoft
he printing and typesetting
industry. Loremipsumis simply
dummytext ofthe printing and
typesetting industry.

Use case for this regex is to format a long string into readable text with max line length enforced and lines starting with a character and not a whitespace.

Optional requirement: When after initial posting I added that example in Edit 1, I also added an optional requirement for adding a dash '-' character at start of the next line, if a word was cut at max line length. I'm removing that from the example now and adding it as a separate optional requirement here.

So an optional requirement: If a line is broken mid-word at max-length and not at a whitespace, then a dash should be appended at the end of that line (and not at start of the next line, as I had originally described).

Loremipsumissimplydummytextoft-  # Line length 30+1 char with an appended a dash
he printing and typesetting     # Cut before the word at otherwise 30 char
industry.

Answer 1

You may use

 var s = "Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry. Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry."; var regex = /\\s*(?:(\\S{30})|([\\s\\S]{1,30})(?!\\S))/g; console.log( s.replace(regex, function($0,$1,$2) { return $1 ? $1 + "-\\n" : $2 + "\\n"; } ) )

Details

• \\s* - 0 or more whitespace chars.
• (?: - start of the non-capturing group:
  • (\\S{30}) - Group 1 (referred to with the $1 variable in the callback method): thirty ( n ) non-whitespace chars
  • | - or
  • ([\\s\\S]{1,30})(?!\\S)) - Group 2 (referred to with the $2 variable in the callback method): any one to thirty ( n ) chars, as many as possible, but not immediately followed with a non-whitespace char.

The function($0,$1,$2) { return $1 ? $1 + "-\\n" : $2 + "\\n"; } function($0,$1,$2) { return $1 ? $1 + "-\\n" : $2 + "\\n"; } function($0,$1,$2) { return $1 ? $1 + "-\\n" : $2 + "\\n"; } part means that if Group 1 matched (that is, we matched a very long word that is cut into two parts), we replace the match with Group 1 value + hyphen and a newline. Else, if Group 2 matches, we replace with Group 2 value + a newline.

ES6+ compliant code snippet :

 const text = "Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry. Loremipsumissimplydummytextofthe printing and typesetting industry. Loremipsumis simply dummytext ofthe printing and typesetting industry."; const lineMaxLen = 30; const wsLookup = 15; // Look backwards n characters for a whitespace const regex = new RegExp(String.raw`\\s*(?:(\\S{${lineMaxLen}})|([\\s\\S]{${lineMaxLen - wsLookup},${lineMaxLen}})(?!\\S))`, 'g'); console.log( text.replace(regex, (_, x, y) => x ? `${x}-\\n` : `${y}\\n`) );

Answer 2

Final answer:

(\\S[\\s\\S]{1,30}$|\\S[\\s\\S]{1,29}(?:\\s+)|\\S{30})

evolution:

1. ([\\s\\S]{1,15}(?!\\S)|\\S{15,})

you just have to modify the answer in the link by an 'or' statement that adds your additional requirement: |\\S{15,}

2. responding to your edits, here is my modified regex: ([\\s\\S]{1,15}(?!\\S)|\\S{15})

you can replace the 15s with 30 or the character cutoff of your choice

3. adjusting for your further clarifications: (\\S[\\s\\S]{1,14}(?:\\s*)|\\S{15})

Now the string has to start with a none-whitespace and it matches but does not capture additional white space after the first 15 characters. Again you need to change the 15 and the 14 to the lengths you want.

4. (\\S[\\s\\S]{1,30}$|\\S[\\s\\S]{1,29}(?:\\s+)|\\S{30}) Adding another condition in the multiple 'or' statement at the beginning which captures the end of the string if it ends in a none-whitespace character. If it ended in a whitespace character, the second part of the 'or' statement captures it.