UNIX for Dummies Questions & Answers

I'm using sh on hp-ux and want to find / print a line that matches 132.101- and the next line, too. grep -A isn't supported on hp-ux, so I'm trying awk and sed.

The code below works but only prints the first occurence. I need all matches from the file.

Code:

awk '/132.101-/{f=2;print;next}f{print;exit}' INFILE1 >outfile1

I've also tried sed but this cannot be parsed, my shell tells me.

Code:

sed -n -e '/132.101-/{N;p}' INFILE1 >outfile1

Thanks for your help!

Code:

awk '/132\.101-/{print;getline;print}' INFILE1 > outfile1

Following are the issues with your awk:

Code:

awk '/132.101-/{f=2;print;next}f{print;exit}' INFILE1 >outfile1

1) The regex is not correct (although it will work). . is a regex metacharacter which matches any character. So, to match a literal ., you'll need to escape the period.

2) The exit statement will make awk stop after it has printed out the first line after the one containing "132.101-". Hence, you are getting only 2 lines.

If you want to use your awk, you'll need to modify it a bit:

Code:

awk '/132.101-/{f=2;print;next}f{print;f=0}' INFILE1 >outfile1

That worked! Thanks a bunch. Never thought that adding something straightforward like ;getline; would be parsed by hp-ux.

I also changed my pattern a bit to find another potential match:

Code:

awk '/132\.10[1-2]-/{print;getline;print}' INFILE1 >outfile1

---------- Post updated at 04:16 PM ---------- Previous update was at 02:25 PM ----------

Just for argument's sake, how would I do the opposite -- pull the line matching the pattern and the previous line? Thanks.

---------- Post updated at 04:42 PM ---------- Previous update was at 04:16 PM ----------

I found an answer after some more digging:

Code:

awk '/132\.101-/{print x;print};{x=$0}' INFILE1 >outfile1