UNIX for Dummies Questions & Answers

I have a slew of files for historical reference.

Each file has a unique line in the file, such as "Today's datetime is:"

Each file also contains a unique set of characters on line X, such as "DMX Info." This line number will be different each time a new file is generated. So, for one file, it might be line 100, whereas the next file might be line 120.

The important info I'd like is always two lines below line X above, but not unique enough to simply grep on by itself. It's called "High Value" and there are several lines like this in each file.

QUESTION:
How do I grep for "Today's datetime is" AND two lines below line X? I don't care if line X is printed. It's the line two lines below that I'm looking for.

THANKS!

not the cleanest solution but:

Code:

awk '/Today's datetime is/{print}/DMX Info/{print t}!/DMX Info/{t=s;s=$0}' file

should work

Thanks for that! Looks clean enough, but appears to be grabbing two lines before line X. I'm looking for two lines after line X. Any insight?

Thanks again!

Try this:

Code:

awk '/Today.s datetime is/;/DMX Info/{getline;print;getline;print;exit}'

Regards

That almost works, but it appears to only scan one file, whereas I'm trying to scan hundreds. Also, anyway to omit printing the first line after line X, and only print the second.

This is very helpful, so thanks guys (or gals).

This should omit the first line after x and scan all files in the current directory:

Code:

awk '/Today.s datetime is/;/DMX Info/{getline;getline;print}' *

There we go. That works, but it appears the 'exit' won't look in all files. For some reason it only scans one. If I omit the 'exit', it works perfectly.

Thanks. Shell scripting is not my strong point, but I need to get moving on it. Thanks again!