Delete regex and 2 lines after that 1

[dstxaix]

I know only a litte bit of scripting and sed and started using awk recently.

I have a scenario where if the first 4 characters of a line are "PASS" I need to delete that line and the 2 lines below that from the file. I was looking at sed and awk one liners page and nothing really suited me for this requirement however they helped me with some of teh other reqs.

I tried

awk '/^PASS/ { if (substr($0, 0, 4) == "PASS") ;else print }' testfile

and

awk '/^PASS/{print x};{x=$0}'

They dont seem to work.

Sample data

TEST
PASS1
ABC
DEF
GHI
PASS2
JKL

MNO

PASS3
PQR

XYZ
123

 

[PHV]

Perhaps this ?

awk '/^PASS/{getline;getline;next}1' testfile

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

 

[dstxaix]

PHV - That is perfect. I gave you a star.