Getting the last match in a file using grep

2 Things:

• As stated by @Rory, you need the -o option, so only the match are printed (instead of whole line)
• In addition, you neet the -P option, to use Perl regular expressions, which include useful elements like Look ahead (?= ) and Look behind (?<= ), those look for parts, but don't actually match and print them.

If you want only the part inside the parenthesis to be matched, do the following:

grep -oP '(?<=\/\()\w(?=\).+\/)' myfile.txt

If the file contains the sting /(a)5667/, grep will print 'a', because:

• /( are found by \/\(, but because they are in a look-behind (?<= ) they are not reported
• a is matched by \w and is thus printed (because of -o )
• )5667/ are found by \).+\/, but because they are in a look-ahead (?= ) they are not reported

grep SESSNUM=4437 *

Grep normally returns thing in order. Are you sure you are getting the results from the correct log file? Does it work correctly if you specify the logfile you want instead of just *?

Correction! There should only be one file in the directory that contains the session number, but I don't necessarily know which one when I'm grepping.

I would suggest you use the -H option of grep, so you can be sure you are getting the content from only one file. This should be the default behavior, but what you posted didn't include the filenames, unless you stripped those out.

   -H, --with-filename
          Print the file name for each match.  This is  the  default  when
          there is more than one file to search.