UNIX for Dummies Questions & Answers

Hi All,

I have a csv file like the following:

Code:

"ABCD2","EFGH2","XXXX","1"
"ABCD2","EFGH2","XXXX","2"

I want to grep out the row which contains the value of 2 within the 4th column, so then i can use the extracted record to cut up and store into numerous variables.

Obviously when trying cat csvfile.csv | grep 2 it greps both lines, as columns 1 and 2 contain the value of 2 also. I have also tried playing about with grep -x but still to no avail.

Any help would be greatly appreciated.

Cheers

Code:

awk -F ','  ' $4~/"2"/ {for (i=1; i<=NF; i++) {print $i} } {next}'  filename.csv > newfile

This gives you the line chopped up into fields in a file. Play with it until you get what you want.

Just had a little play around with what you suggested, and found that the following works:

Code:

awk -F ','  '$4~/"2"/' <filename.csv>

The above returns only the record i require, and does not require a new file been created etc. I had played about with awk -F , '$4 == 2' but that didnt work, however using the "~" and "/" does work.

Thankyou!

@RichZR: If the file contains only 4 columns, then you could also do this:

Code:

grep '2"$' input.txt