How to use regex match end of line in Windows

I have a .txt file created in Windows and now should be edited in Linux. I want to match the end of a line with grep. Let's say the content of the line I am going to find is "foo bar" in file bar. Then I issue the command grep 'r$' bar, but no output yielded.

Given in Windows a new line consists of '\r\n', diffe开发者_如何学Crent from Linux/Unix a single '\n', I think there must be something subtle related to this. Then I convert the file with dos2unix and voila, it works.

How can I match the content without convert the original file?

Use a pattern which matches both line ends: \r?\n

If your grep supports -P (perl-regexp), then to match a CRLF:

grep -P '\r$' file

or:

grep Ctrl+VCtrl+M file

(Ctrl+VCtrl+M will produce ^M)

Give it a try with AWK:

awk '/r$/' RS='\r\n' bar

or

awk '/r\r$/' bar

I had a similar issue in Windows Subsystem for Linux (WSL), running GNU Bash (v4.3.48) on Ubuntu. None of the suggestions above worked for me, but the following did (thanks to @phuclv for the tip to use \015 instead of Ctrl + V, Ctrl + M).

grep -Prno "TOKEN[^\015\012]*" *

I wanted to match from TOKEN to the end of line and was getting back blanks. So I tried matching all the non carriage returns (\015), and non newlines (\012) after TOKEN.