[SOLVED] Grep last match until the end

Issue

I've looked around StackExchange sites but I haven't found anything that's quite what I'm looking for. Here are two use cases of grep:

• Printing items before/after a match
• Print a certain match

I'm trying to parse a log file, and I want to return the last error in the log which is, predictably, at the end of the file. However, sometimes the errors are multiple lines. The answers for 'how to grep the last match' all involve either tail or head, and only work with a single line.

In my case, I want to simply return everything in the file, starting with the last match. Typically, this won't be any more than 10-15 lines maximum, so a grep -A 15 does the trick there. But, I still need to only get the last one of these, so that alone doesn't produce the right output.

The naive approach is to use a two-part match, to first get what the last match is and then everything after that. This won't work for me, because I can't guarantee that the last match is unique.

Is it possible to do this with grep somehow, or would there be better tools for this?

Solution

There is a way to get sed to do this but I can't remember.

If you are open to using a combination of commands here is something that might work:

# Get the line number of teh last match

LNO=$( grep -n 'the error' the_file | tail -1 | cut -d":" -f1 )

# Now use sed to print all lines from that point:

sed -n "$LNO,\$p" the_file

Answered By - TenG