How to print the line numbers with corresponding line that matches a pattern using shell command?

Question

File,

.false alarm is bla no. 11
.no alarm generated is bla
.application will be killed is bla wall command
.doc file is document file with .doc extension
.no authority for this selection bla

Now what I want in my output file to print the lines which contains only the word bla (Or it may be number also)

Output will be like this,

1 .false alarm is bla no. 11
2 .no alarm generated is bla
3 .application will be killed is bla wall command
5 .no authority for this selection bla

Solution

Many tools can be handy:

• -n of grep is exactly what you are looking for.

  grep -n 'bla' file
  

• alternatively awk:

  awk '/bla/{print NR":"$0}' file
  

• alternatively perl:

  perl -ne 'print $.,":",$_ if /bla/' file
  

• alternatively sed:

  sed '/bla/!d;=' file |sed 'N;s/\n/:/'
  

Comments

[Admin]

awk 'i++ {if($1~/RRBS/) print i}' ../../bak/bak.db

[Nainita]

:-) yes... got the desired output. @jimmij

[Admin]

sed can also be sed -n '/bla/!d;=;p' file