Shell Programming and Scripting

I am trying to extract a table of data (mysql query output) from a log file. I need to print everything below the header and not past the end of the table. I have spent many hours searching with little progress. I am matching the regexp

Code:

+-\{99\}

with no problem. I just can't figure out how to print just between the second and third matches. I would like to do this with sed, because I am already useing sed a couple of other places in the script.
Thanks for your help.

Please post a representative sample of the input and the corresponding desired output.
Also, do not try to oversimplify things; else you might end up with a long thread before getting any proper solution.

Otherwise try:

Code:

awk '/\+-{99}\+/{p++; next}p==2' file

--
With GNU awk < 4.0 Try awk --posix
or try:

Code:

awk '/\+-+\+/{p++; next}p==2' file

The last line worked just fine. My awk version is mawk 1.3.3 Nov 1996. Please explain the the search pattern used here. I have just about worn the print off the pages of my O'Reilly book.

Hi the first line does not work with mawk since mawk does not support the braced repetition operator {..}. The last line uses the simple repetition operator + which means 1 or more. So this extended regular expression means a plus-sign \+ followed by one or more minus-signs -+ followed by a plus-sign \+..

Thanks again; much appreciated.