UNIX for Dummies Questions & Answers

Hi all,

I have a TAB separated file like this:

sample.rpt:

54 67 common/bin/my/home {{bla bla bla}} {bla bla} Replace Me
89 75 bad/rainy/day/out {{ some bla} } {some bla} Dontreplace Me

......
......

I wish to do a regexp match on the 3rd item, say "common/bin/my/home" and if there is a match i want to replace the 6th item "Replace Me" by my own text like "happy Home"

Additionally my regexp will be a shell variable, so I need a command which accepts regexp as variable

The items may have spaces or backslashes and the file is Tab separated.

Is there a simple one liner using sed or awk which can perform this over entire file in one command ?


Thanks in advance

Try...

Code:

awk 'BEGIN{FS=OFS="\t"}$3~s{$6=r}1' s='common/bin/my/home' r='happy Home' sample.rpt

Thanks for soln. I figured out one way to d it myself:

awk -F "\t" -v drv=$mod_pat '$3 ~ drv { OFS= "\t"; $6 = " happy_home"; print;}' sample

---------- Post updated at 06:34 PM ---------- Previous update was at 06:19 PM ----------

I got caught in the else part of condition:

Below I am checking if field 3 matches the pattern specified by drv do whatever next in braces {}

awk -F "\t" -v drv=$mod_pat '$3 ~ drv { OFS= "\t"; $6 = " happy_home"; print;}' sample


The problem now is I want to just print the line untouched if the pattern is not matched. Is this possible in the same command ?
I tried:

awk -F "\t" -v drv=$mod_pat ' if ($3 ~ drv) { OFS= "\t"; $6 = " happy_home"; print;} else {print;}' sample.rpt


But its throwing parse error. Probably some problem with awk syntax.


Any help will be appreciated.

Thanks

---------- Post updated at 09:00 PM ---------- Previous update was at 06:34 PM ----------

Anyhelp ? is this thread closed ??

---------- Post updated at 09:39 PM ---------- Previous update was at 09:00 PM ----------

Thanx anyways, got it:

This works:

awk -F "\t" -v drv=$mod_pat '{ if ($3 ~ drv) { OFS= "\t"; $6 = " happy_home"; print;} else {print;} }' sample.rpt