Passing output from one command as argument to another

Question

Hello,

I have this for:

for i in `ls -1 access.log*`; do tail $i |awk {'print $4'} |cut -d: -f 1 |grep - $i > $i.output; done

ls will give access.log, access.log.1, access.log.2 etc.
tail will give me the last line of each file, which looks like: 192.168.1.23 - - [08/Oct/2010:14:05:04 +0300] etc. etc. etc
awk+cut will extract the date (08/Oct/2010 - but different in each access.log), which will allow me to grep for it and redirect the output to a separate file.

But I cannot seem to pass the output of awk+cut to grep.

The reason for all this is that those access logs include lines with more than one date (06/Oct, 07/Oct, 08/Oct) and I just need the lines with the most recent date.

How can I achieve this?

Thank you.

Answer 1

Umm... Use xargs or backticks.

man xargs

or http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html , section 3.4.5. Command substitution

Answer 2

why don't you break it up into steps??

for file in *access.log
do
  what=$(tail "$i" |awk {'print $4'} |cut -d: -f 1)
  grep "$what" "$file" >> output
done

Answer 3

As a sidenote, tail displays the last 10 lines.

A possible solution would be to grepthis way:

for i in `ls -lf access.log*`; do grep $(tail $i |awk {'print $4'} |cut -d: -f 1| sed 's/\[/\\[/') $i > $i.output; done

Answer 4

You shouldn't use ls that way. Also, ls -l gives you information you don't need. The -f option to grep will allow you to pipe the pattern to grep. Always quote variables that contain filenames.

for i in access.log*; do awk 'END {sub(":.*","",$4); print substr($4,2)}' "$i" | grep -f - $i > "$i.output"; done

I also eliminated tail and cut since AWK can do their jobs.