Hi, I am a newer for Bourne shell, I have a log file for a unix system, it some thing like:
35.75.253.207 - - [17/Aug/2003:12:42:37 +1000] "GET /products/orgonizer/title.png HTTP/1.1" 200 1555 "-" "Mozilla 1.4"
Now I want to list 10 most requested files in the log in a descending order, it something like
157 /index.html
129 /title.gif
...
How should I count the duplicate file and display the number of requested in front of the file requested, like above, is there command avaliable to achieve that? I just know sort -nr is displaying the above in a descending order.
Thanks for help

Use 'uniq' with the -c option.

uniq -c yourFile

Pipe into sort if you want to order it...

uniq -c yourFile | sort -r

Pipe into head if you only want n lines...

uniq -c yourFile | sort -r | head -n # where n is the number of lines

If for example the 10th and 11th lines had the same count and you wanted to return both, you would have to do something a little more sophisticated...

uniq -c yourFile |
sort -r |
awk -v numRows=10 'NR>numRows && $1<lastCount {exit}{print; lastCount=$1}'


HTH

Thanks, that is pretty helpful!