Here is my input file:
/home/sites/default/files/Maple%20board02%2019%2013%20.pdf
/home/sites/default/files/paintgrade/side-view.jpg
/home/sites/default/files/paintgrade/steps_2.jpg
/home/sites/default/files/Front%20sill-photo1.gif
/home/sites/default/files/Rear%20steps%20Feb.%209.2011.pdf
Here is my grep/awk statement:
grep /files/ 404s.txt | awk -F '/' '{print $6"/"$7}'
The output from that statement is:
Maple%20board02%2019%2013%20.pdf/
paintgrade/side-view.jpg
paintgrade/steps_2.jpg
Front%20sill-photo1.gif/
Rear%20steps%20Feb.%209.2011.pdf/
(See the trailing slash? Sometimes it's there, sometimes not. Sometimes there isn't a subdirectory and my awk statement prints a "/" none-the-less.)
I saw on another post how to remove a trailing slash, but I am not sure how to apply it here. The post said that ${1%/} will give you a string without a trailing slash. The post is here. The highest voted answer is: target=${1%/}
I'd like to add something to my grep/awk statement to have the output be:
Maple%20board02%2019%2013%20.pdf
paintgrade/side-view.jpg
paintgrade/steps_2.jpg
Front%20sill-photo1.gif
Rear%20steps%20Feb.%209.2011.pdf
What can I add to my original statement to have the output be as above? Maybe cut could help or my awk could be adjusted not to print a trailing "/"?
To remove just that prefix something like this will work (substitute out the string and then use a truth-y pattern to get the default print action).
awk '{sub("^/home/sites/default/files/", "")}7'
If you need to remove X fields from the start of the line then using cut generally makes that simpler than awk.
cut -d/ -f6-