Just trying to cut any characters after the 7th on each line of a text.

I initially tried this:

sed 's/(.{7}).*/\1/' TestText.txt 

But got this response:

sed: 1: "s/(.{7}).*/\1/": \1 not defined in the RE

Then tried this:

$ sed 's/./&#/7' TestText.txt 
$ sed -e 's/#.*//' TestText.txt 

But all it did was remove the #. I just want to remove any character after the 7th character.


Use sed with -r:

sed -r 's/(.{7}).*/\1/' file

Or escape the brackets:

sed 's/\(.\{7\}\).*/\1/' file

Alternatively, you can also use grep (-E activates extended regular expressions, -o prints only the matching pattern):

grep -oE '^.{7}' file

With awk:

awk '{print substr($0,1,7)}' file

And of course cut is just made for this kind of job:

cut -c1-7 file
  • 1
    You should use -E instead of -r. -E is supported by GNU and some BSDs and is going to be in the next POSIX spec. (and -r doesn't make sense) – Stéphane Chazelas Dec 15 '15 at 13:39
  • 1
    Note that on GNU systems, cut -c doesn't support multi-byte characters, using sed is a work around. – Stéphane Chazelas Dec 15 '15 at 13:42
  • 1
    grep -oE '^.{7}' (a GNUism) will remove the lines that have fewer than 7 characters. – Stéphane Chazelas Dec 15 '15 at 13:43
  • I used the cut method and that worked. Cheers! – Phil Dec 15 '15 at 13:47
  • 1
    @StéphaneChazelas Regarding -E and -r are you talking about sed ? – 123 Dec 15 '15 at 14:32

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