How do I grep lines ending on a specific expression

Question

I have a text file with 4 sentences and I want to grep the ones ending in "me", however when I do grep "me$" it doesn't work.

Am I missing something?

(I wanted the result to be the 1st, 2nd and 4th line).

Edit: Replaced image with text.

$ cat sm.txt
somebody once told me

somebodyoncetoldme

the world was

gonna roll me
$ grep "me$" sm.txt
$

Answer 1

In comments you say that file identifies the text file as an ASCII text file with CRLF line endings. That's another way of saying it's a DOS text file, possibly generated with an editor on a Windows system.

A DOS text file has, in comparison to a Unix text file, an extra carriage return character at the end of each line, before the newline. This is why your regular expression does not match, no line actually ends with me, they end with me\r, where \r is a common way of writing a carriage return character.

The remedy for this is to convert your text file to a Unix text file using e.g. the dos2unix conversion tool, before using it on a Unix system.

Answer 2

In itself grep "me$" definitely is correct. I suspect your lines end with a few additional characters that are hard to see, like tabs and spaces.

Try one of these:

# \t means tab
# [\t ] means a tab or a space
# [\t ]* means any number (including zero) of tabs and spaces.
#
grep "me[\t ]*$" sm.txt

Or:

# [[:space:]] means any white space. I don't know if there are
# more than just space and tabs. This works with GNU grep only.
#
grep "me[[:space:]]*$" sm.txt