How to print one line before and one line after certain pattern?

[whatisthis987]

Hi guys,
I would like to print the line before and the line after "JohnDoe" is found. For example,

===Input text file===
line1
line2
JohnDoe blah blah blah
line4
line5

===Desired output file===
line2
line4

Can someone suggest a good way of doing this?

 

[columb]

If you have a suitable version of grep then

grep -A 1 -B 1 JohnDoe InputFile | grep -v JohnDoe

On the internet no one knows you're a dog

Columb Healy

 

[LKBrwnDBA]

On Linux:

man grep, look at the -A and -B options.


----------------------------------------------------------------------------
The person who says it can't be done should not interrupt the person doing it. -- Chinese proverb

 

[whatisthis987]

Thank you guys. I never knew the usage of -A and -B. This is perfect.

 

[PHV]

Another way:
awk '/JohnDoe/{print x;getline;print}{x=$0}' /path/to/input

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[Annihilannic]

Or grep -C1 is equivalent to -A1 and -B1. You only need to specify before and after separately if they are different.

Annihilannic.