[SOLVED] Pattern Searching

whocares357 · 11-02-2010, 02:56 PM

I have a file with a random word on each line (3k+ lines).

How can I get the lines with only five characters? I tried using grep '.....' file | more, but it returns all the words (even those less than 5 characters).

Edit: I also tried grep '.{5}' file | more but it doesn't show anything. And grep '.\{5\}' file | more returns all lines with four or more characters (I'm really confused about why it's doing this).

sycamorex · 11-02-2010, 03:08 PM

It's not the most elegant solution but I'd do:

Code:

sed -n '/^.....$/p' file

whocares357 · 11-02-2010, 03:13 PM

Quote:

Originally Posted by sycamorex

It's not the most elegant solution but I'd do:

Code:

sed -n '/^.....$/p' file

Hm, I must have some hidden characters in the text file because my output consists of all four character lines. Is there any way to reveal hidden characters (in vi or cat)? Nevermind: set list in vi.

Edit: Thank you very much. I actually simplified the expression with '/^.\{5\}$/p'.

sycamorex · 11-02-2010, 03:41 PM

Excellent. That's much better.

crts · 11-02-2010, 05:06 PM

Quote:

Originally Posted by whocares357

Hm, I must have some hidden characters in the text file because my output consists of all four character lines. Is there any way to reveal hidden characters (in vi or cat)? Nevermind: set list in vi.

Edit: Thank you very much. I actually simplified the expression with '/^.\{5\}$/p'.

Hi,

to further "simplify" you could use sed's -r option:

Code:

sed -r -n '/^.{5}/p' infile

This way you do not have to escape the curly braces {}. Ok, it is not really simplifying but readability improves - especially if you have to use a lot of {()} ...