I am attempting to search through several lines of text with the following reg expression using grep:

Code:
grep -o '9[0-9] [8-9][0-9] 9[0-9].*\{40\}' sample.txt
Works but grep does not always return the following 40 characters. Instead it only returns up to 40 characters to the EOL which ever comes first.

What I am looking to do is get the following 40 characters whether there is an EOL within the following 40 characters or not.

Rather not show with what I have in sample.txt but here is the idea:

sample.txt
Code:
90 80 90 hello there
good bye
returns....

Code:
90 80 90 hello there good

Well I am not sure I understand how that regex works at all? After the number section you say, return zero or more of any character and then you want 40 lots of the previous expression???

As to answer your question, if newlines are allowed in the solution then you will probably need to include them in your regex.

Native grep works on single lines and cannot do multi-line matches. There is a somewhat experimental "-P" option to use Perl regular expressions, which would support that. Here is a link to a discussion with an example of multi-line matches using "grep -P".

-a1 did the job. Thanks for your help

And if the 40 characters were to go over more than 1 extra line? How will you know to make -A2 or more?

In this situation that would not be the case.

Well if happy with your solution, please mark the question as SOLVED