Shell Programming and Scripting

hi..i have a file like the following

Application handle = 2604
Application status = UOW Waiting
Application handle = 2384
Application status = UOW Waiting
Application handle = 2383
Application status = UOW Waiting
Application handle = 1679
Application status = Lock-wait
Application handle = 2248
Application status = UOW Waiting
Application handle = 2236
Application status = UOW Waiting
Application handle = 2382
Application status = UOW Waiting
Application handle = 2863
Application status = Lock-wait
Application handle = 2381
Application status = UOW Waiting
Application handle = 2380
Application status = UOW Waiting


here i need just only the lines with
Application handle = 1679
Application status = Lock-wait
Application handle = 2863
Application status = Lock-wait

ie the line with Application status = Lock-wait and the line above it

any help is appreciated.

If your grep groks the -B option,

Code:

grep -B1 Lock-wait file

Or hack your own little awk script:

Code:

awk '/Lock-wait/ { print prev; print } { prev = $0 }' file

Below will give you the line with Lock-wait and the next line. have to check to get the previous line.

sed -n '/Lock-wait/{N;p;}' <FILENAME>

You can do the same in sed as in the awk script: remember the previous line, then if there's a match, print the remembered data and the current line.

Code:

sed -n '/Lock-wait/{x;p;x;p;d;};h' file

thank you very much guys...

oops sorry forgot to ask about printing the below line also along with line above the Lock-wait..

Code:

grep -A 1 -B 1 Lock-wait file

Code:

a=$1
b=$2
cat file | sed -n "/$a/{
p
n
p
}
/$b/{
p
n
p
}"