This isn't at all easy to explain but easy to just show.

I have lines in a file such as:

100Dollars              3              IP  
200Dollars              3              IP
300Dollars              4              IP

I need to grep for lines that have no '3' in the second column. I tried the following:

egrep -v '3' filename

However this does not return the third line due to having a 3 in the first part of it. There's my basic question, if that makes sense.

How do I exclude what is in the first column and only grep for whatever is in the second column?

  • You say "2nd column" then you say "third column"; which is it?
    – Hello71
    Mar 23, 2011 at 21:41
  • 2nd column. Fixed.
    – roger34
    Mar 23, 2011 at 21:50
  • This question has nothing to do with bash. Jun 16, 2014 at 15:57

4 Answers 4


Can't you just do grep -v " 3 ", assuming the columns are delimited by spaces?

  • Gave this a shot, does not work. Opened the file in vim and it looks like it is separated by tabs in the columns, not spaces.
    – roger34
    Mar 23, 2011 at 21:47
  • @roger34: Then use tabs... duh.
    – Hello71
    Mar 23, 2011 at 21:49
  • Did, it didn't work.
    – roger34
    Mar 23, 2011 at 21:53
  • 1
    Try grep -v '[[:space:]]3[[:space:]]' or grep -v '\<3\>'.
    – Mikel
    Mar 23, 2011 at 22:08
  • Mikel the second regex worked. Thanks. Now I have to investigate why it works and learn from it.
    – roger34
    Mar 23, 2011 at 22:12

How about:

awk '$2 != 3' filename

I think you want to step up to awk or grep, which can do columns (among many other things)

gawk '$1 ~ /3/ && $2 !~ /3/{print $0}' < filename

Should do it.

This looks for a 3 in the first column (Columns are numbered starting by 1 in awk, $0 is the whole line) and not a 3 in the second column, and if so print the whole line ($0)


You can use:

grep -v -P '\t3\t' filename

-P is a perl-style regular expression matcher.

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