Shell Programming and Scripting

Here Is a problem I am facing with awk.

Query --> I want to search for a string in a file and print next 15 lines below the matched string.

1.We do not have GNU grep so cannot use grep -A or grep -B commands.
2. Instead of passing the search pattern as a string to awk. I want the awk to read the pattern from a file.

This is the command which I am using:

Code:

cat fulldatafile.txt | awk 'c-->0;/'`cat IDnumber.txt`'/{c=17}'

So what I did:
cat the full data file (file which contains information associated with ID) and then using the awk command I want to search for the only single ID, the ID is in the IDnumberfile which is updated by an external program sometimes. Kindly, help me in finding out how to read a file to get the pattern.

With whatever little awk knowledge I have, this is what I could assimilate into a solution:

Code:

awk '/pattern/ {c=0; f=1; print; next} f==1 && c<=15 {print; c++} c==15 {exit}' inputfile

After a bit of googling, I found a solution in this link.

Code:

awk '/pattern/{c=15;{print}next}c-->0' inputfile

Try this too -

Code:

awk '/pattern/ {for(i=0;i<15;i++) {print;if(!getline) exit;} }' inputFile

Code:

[ -r IDnumber.txt ] && awk 'FNR==NR{patt=$0;next} index($0,patt){for(i=1;i<=15 && getline;i++) print}' IDnumber.txt fulldatafile.txt

Thank You Guys, let me try these solutions out and I will update you shortly.