查找某些/特定的换行符，同时忽略其他换行符

Question

I have some SRT data that is coming back with \\r and \\n tags as line breaks in the middle of each sentence. How do I find only those \\r and \\n tags in the middle of the text/sentences and NOT the other ones that signify other line-breaks.

Example source:

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.

Desired output:

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange.

I am absolute crap at regex, but my best guess (to no avail) was something like

var reg = /[\\d\\r][a-zA-z0-9\\s+]+[\\r]/

And then split() on that to remove any \\r in the middle of one of the values. I am sure that is not even close to the right way so...stackoverflow!! :)

Answer 1

This will match the line breaks you want to get rid of, capturing the character before and after it, to put those two back in place around a space:

var regex = /([a-z,.;:'"])(?:\r\n?|\n)([a-z])/gi;
str = str.replace(regex, '$1 $2');

Some things about the regular expression. I used the modifiers i and g to make it case-insensitive and to find all line breaks in your string instead of stopping after the first one. Also, it assumes that removable line breaks can occur after a letter, comma, period, semicolon, colon or single or double quotes and before another letter. As @nnnnnn mentioned in a comment above, this will not cover all possible sentences, but it should at least not choke on most punctuation. The line break does have to be a single line break, but it is platform-independent (can be either \\r , \\n or \\r\\b ). I capture both the character before the line break and the letter after the line break (with parentheses), so I can access them in the replacement string with $1 and $2 . That is basically all there is to it.

Answer 2

This regex should do the trick:

/(\\d+\\r\\d{2}:\\d{2}:\\d{2},\\d{3} --> \\d{2}:\\d{2}:\\d{2},\\d{3}\\r)([^\\r]+)\\r([^\\r]+)(\\r|$)/g

To make this work with more lines (has to be a set number) then just add more ([^\\r]+)\\r 's. (Remember to also add $ 's to the match replace as so (with 3 lines): '$1$2 $3 $4\\r' ).

Usage

mystring = mystring.replace(/(\d+\r\d{2}:\d{2}:\d{2},\d{3} --> \d{2}:\d{2}:\d{2},\d{3}\r)([^\r]+)\r([^\r]+)(\r|$)/g, '$1$2 $3\r');

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Limitations

• If there is more than 2 lines of text this won't work.

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Example 1

Works fine!

Input:

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.

Output:

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange

---

Example 2

Doesn't work; more than 2 lines

Input:

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,
and they just talk

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.
Except for Maria who dyes it pink.

Output:

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,
and they just talk

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange.
Except for Maria who dyes it pink.