How to grep for a line that has the same number you both need and don't need

Question

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?

Answer 1

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

Answer 2

How about:

awk '$2 != 3' filename

Answer 3

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)

Answer 4

You can use:

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

-P is a perl-style regular expression matcher.