Echo final value of long line
 

Hi guys

I am writing a command to return the last value entry within a log file:
This works fine and prints what I expect

However I only want to echo the very last value of this line which in this case 2419, How is this possible?

Thanks in advance

One simple way would be with cut command.
If number of items can vary from lines to lines, I can suggest you awk because it's easy to access the last item with it, otherwise you need to reverse the line with rev and cut the 1st item and re-rev the line).
Can you try something and tell us?

Pipe your output to:

Fantastic it worked thanks guys :D... Can you just explain what sed -e 's|^.*\s\(.*\)$|\1|' is doing?

The ^ just marks the beginning of a line. .* says match any character zero or more times. \s says match a white space. \( marks the beginning of a lookback section. Again .* says match anything zero or more times. \) marks the end of the lookback. $ marks the end of the line. Sed regex is greedy. So the .* grabs everything it can, but \s.* says to keep the last whitespace delimited bunch of characters separate. That is then the only thing we print out with the \1 back reference.

Hi guys

Something I just thought, the file name will change depending on the date etc. localhost_access_log.2018-10-17.txt and now today it is localhost_access_log.2018-10-18.txt.

How in a script do we determine to search for the log file by date? Would we be using the date command?

Something along the lines of

Thanks in advance

Cheers

Maybe you could just take the "last" one (see warning below) as input with something like:
EDIT: option -c with ls instead of -t. /!\ Since -c option sorts by ctime (there doesn't seem to be an easy and universal way to sort by creation time), it works as long as information about the log files are not modified otherwise your sort output could be different from what you would have expected...

Thank you all so much

New requirement from the team!

They are now saying they want the lines which include 10.218.182.157 - - [18/Oct/2018:09:42:27 +0000] "POST /HvpWorker/hvpCommand HTTP/1.1" 200 172 1
to be removed.

I have tried using but this doesn't seem to work?

I believe they want to see the first 5 lines of

| head -n 5 |

But this comes back blank. I'm assuming the first sed will need to come before the second head?

should extract all of those lines. If you just want the first five, pipe that to 'head -5'

Since OP says "They are now saying they want the lines which include 10.218.182.157 - - [18/Oct/2018:09:42:27 +0000] "POST /HvpWorker/hvpCommand HTTP/1.1" 200 172 1 to be removed" I've understood the problem differently. My answer would be more like this:

Hi bradvan

Sorry probably didn't explain that very well.

This command worked fine and got what I needed. Then adding the gave me just the numbers.

However now I have been told we don't need the following lines(In Bold)

So I need the first 5 lines within that log file which doesn't include the single figure if that's possible. In the log it seems to be every other line?

Thanks in advance

I've worked out how to do it as such

I was wondering if instead of greping for JobId could we remove every other line or is that not possible?

Ok, I got it: