CODEWITHSUNDEEP
awk
judi
judi

Reputation: 99

awk script to eliminate special characters

My output looks like below 

judi# *diff -C 0 $* mountYday mountTday
*** mountYday   Sat Jun 13 02:57:09 2015
--- mountTday   Sat Jun 13 02:59:49 2015
***************
*** 20 ****
- /test on /dev/vx/dsk/test/test read/write/setuid/devices/intr/largefiles/logging/xattr/onerror=panic/dev=48986a5 on Wed Apr 22 22:28:04 2015
--- 19 ----
judi#

I need to get only the line starts with single "-". The output should be as 

- /test on /dev/vx/dsk/test/test read/write/setuid/devices/intr/largefiles/logging/xattr/onerror=panic/dev=48986a5 on Wed Apr 22 22:28:04 2015

I am able to eliminate the first two lines using awk:

*# diff -C 0 $* mountYday mountTday | awk '! /mount/'*

***************
*** 20 ****
- /ora03/oradata1 on /dev/vx/dsk/GRPA056L/ora03-oradata1 read/write/setuid/devices/intr/largefiles/logging/xattr/onerror=panic/dev=48986a5 on Wed Apr 22 22:28:04 2015
--- 19 ----

How do I remove the lines starts with *** and --- (exactly 3 *** and ---) and the *************** (15 Characters)? I am trying with awk in single command.

Upvotes: 1

Views: 59

Answers (1)

Rob Latham
Rob Latham

Reputation: 5223

Instead of removing characters you don't want, how about a command to find the character you do want? The line you want is "start of line, followed by a single -, followed by space". 

$  diff -C 0 $* mountYday mountTday | awk /- / {print $0}

This sort of pattern extraction is more suited to grep:

$  diff -C 0 $* mountYday mountTday | grep '^- '

Upvotes: 2

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