sed print all lines between second and third identical lines

Shell Programming and Scripting
1

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

+-\{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.

2

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.

3

Otherwise try:

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

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

awk '/\+-+\+/{p++; next}p==2' file
4

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.

5

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 \+ ..

6

Thanks again; much appreciated.