Print nth column of pattern

cat file_name | awk {print $n} # where N is the column no. and tab is field seperator

cat SO_02052012_2.csv | awk -F; {print $n} # where N is is column no. and semicolon is field seperator

awk {print $n} < filename

where n is the field number

Question is without using awk commands, any idea ??????

perl -ane print "$F[0]"; < test.txt

You can use this I hope , replace the zero with the column number you want to print.Normally people use cut and awk command for extracting column.

i think by modifying the IFS, we can achieve this as follows, i hope this helps.

#take for example there is an entry in /etc/password with this info
santarvedi:x:500:500:santarvedi:/home/santarvedi:/bin/bash

#let me extract my 6th column, ie, my home directory (/home/santarvedi) using the following mode.

santarvedi]$ bk_IFS=$IFS
santarvedi]$ IFS=:
santarvedi]$ set -- $(getent passwd santarvedi)
santarvedi]$ echo $6
santarvedi]$ IFS="$bk_IFS"