正则表达式从行尾删除回车符（如果存在）

Question

The data looks like this:

0.43,0.57,0.71,0.29,0.71,1.00,0.57,1.00\r
0.43,0.57,0.71,0.29,0.71,1.00,0.57,1.00
0.43,0.57,0.71,0.29,0.71,1.00,0.57,1.00\r

and I want to extract the carriage return characters (marked with \\r in the sample above). I have been trying this using gm and a capture group:

(.*)(?:\\r)$

but this matches only the lines with an \\r . I thought the solution would be to add a ? before the $ but this does not work.

demo: https://regex101.com/r/jArLdS/1

Cheers

Answer 1

If you want to match all lines excluding \\r - a carriage return - use

[^\r\n]+

that is a negated character class ( [^...] ) that matches one or more (due to + quantifier) characters other than \\r and \\n .

Answer 2

Live Demo

Try this regex:

^([^\\r\n]*)(:?\\r)?$

This is assuming you mean excluding the \\r and not extracting \\r because your original regex does that.

Answer 3

Portable solution.

(^.*?)(\\\\r)(\\r\\n|\\n)

That's 3 capture groups, left-to-right, starting at 1 , not 0 .

Replace your string with capture groups 1 and 3, skipping 2.

Capture groups usually syntactically like \\1 or $1 for most languages that use them.