Print lines containing an exact word, not regex

Asked Sep 08 '17 at 14:22

Active Sep 08 '17 at 15:13

Viewed 38 times

I'm writing a bash script and I would like to print all lines containing a word, but only if it is at the end of a line. So given a file like this:

d41d8cd98f00b204e9800998ecf8427e  ./12/2.txt
d41d8cd98f00b204e9800998ecf8427e  ./12/1.txt
717c41ff4049b0e8cbdc7ec7e49ad021  ./1

and the given word equal to "./1" I would like to print only the line:

717c41ff4049b0e8cbdc7ec7e49ad021  ./1

I'd simply use grep for that, but my problem is that the words may contain dots so I need to have the -F option, but then I am unsure how to secure that the printed lines contain the word at the end of the line, as I can't use line anchors. What seems to be working is using grep -Fw, but I am not convinced that this exactly what I'm looking for as -w option means --word-regexp and I don't want regex. Does the-F option disable the regex matching in -w? Is there a better way to achieve my goal?

edited Sep 08 '17 at 15:13

asked Sep 08 '17 at 14:22

madasionka

1

`man grep`, and look at the `-F` (`--fixed-strings`) option. – larsks Sep 08 '17 at 14:25
Thanks @larsks! I edited the post, this is indeed the answer to my previous question, but not exactly what I was looking for as the question was badly asked. – madasionka Sep 08 '17 at 15:14
why not just use 'grep -E "1$" file.txt '. What is the reason you cant use a regex? That would work perfectly according to your example. – Matias Barrios Sep 08 '17 at 19:05
It doesn't work with this example. The second line is also selected. – madasionka Sep 11 '17 at 07:56

Print lines containing an exact word, not regex

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