Print the above and below lines for the grep pattern.

Shell Programming and Scripting
1

Hi,

i would like to get the above and below lines of the grep pattern .
For ex :
file as below:
chk1- aaaa
1-Nov
chk2 -aaaa
##########
chk1-bbbbbb
1-Nov
chk2-bbbbbb
#########

my search pattern is date : 1-Nov

i need the o/p as below
chk1- aaaa
1-Nov
chk2 -aaaa
chk1-bbbbbb
1-Nov
chk2-bbbbbb

Thanks in Advance :slight_smile:

2 
$ grep -B1 -A1 1-Nov inputfile
chk1- aaaa
1-Nov
chk2 -aaaa
--
chk1-bbbbbb
1-Nov
chk2-bbbbbb
3

If your grep doesn't support that option, with awk...

 
awk '/1-Nov/{print prev"\n"$0;getline;print}{prev=$0}' infile
4

Try this,

awk '{if(/1-Nov/){print a"\n"$0;getline;print $0} else {a=$0}}' inputfile
5

got error :

illegal option -A
illegal option -1

---------- Post updated at 11:49 AM ---------- Previous update was at 11:38 AM ----------

its working :slight_smile: .

if i want the below two lines ..
kindly help

6 
awk '/1-Nov/{print prev"\n"$0;getline;print;getline;print}{prev=$0}' infile
7 
grep -C 1 "string" file

To get the same behaviour as grep -A/B/C we need to check if the line exists before printing it:

awk '/string/{if(p)print p;print;if(getline)print}{p=$0}' infile

or

sed -n '/string/{1!x;1!G;$!N;p;};h' infile

Otherwise there is an extra newline or a double last line...