Use grep to display only line numbers

Question

I have a file that may contain incorrect formatting (in this case, the appearance of the \\backslash template). I would like to use grep to only return line numbers where this happens (as in, the match was here, go to line # x and fix it).

However, there seems to be no way to print the line number ( grep -n ), not a match or line.

I can use another regular expression to extract line numbers, but I want grep to not be able to do this myself. grep -no comes closer, I think, but still displays a match.

Answer 1

try the following:

 grep -n "text to find" file.ext |cut -f1 -d: 

Answer 2

If you are open to using AWK:

 awk '/textstring/ {print FNR}' textfile 

In this case, FNR is the line number. AWK is a great tool when you look at grep | cut, or anytime you want to use grep output and manipulate it.

Answer 3

All of these answers require grep to generate all matching lines and then translate them into another program. If your lines are very long, it might be more efficient to use only sed to output line numbers:

 sed -n '/pattern/=' filename 

Answer 4

Bash version

  lineno=$(grep -n "pattern" filename) lineno=${lineno%%:*} 

Answer 5

You will need the second field after the colon, not the first.

grep -n "text to find" file.txt | cut -f2 -d:

Answer 6

To count the number of lines matching a pattern:

 grep -n "Pattern" in_file.ext | wc -l 

To retrieve the mapped pattern

 sed -n '/pattern/p' file.est 

To display the line numbers on which the pattern was mapped

 grep -n "pattern" file.ext | cut -f1 -d: 

Answer 7

I recommend sed and awk answers only to get the line number, instead of using grep to get the entire matching line, and then remove it from the output with cut or another tool. For completeness, you can also use Perl:

 perl -nE '/pattern/ && say $.' filename 

or Ruby:

 ruby -ne 'puts $. if /pattern/' filename 

Answer 8

using grep only:

grep -n "text to find" file.ext | grep -Po '^[^:]+'