I have a file of file paths. The depth of the directories is of various lengths and path names. I'd like to match two directories backwards (two /'s) and delete the match, creating a new file with the results.
eg:
/dir1/dir2/dir3/dir4/dir5/dir6/dir7/output_job3344.xml
/dir1/dir2/dir3/dir4/dir5/otherfile.txt
the result would be:
/dir1/dir2/dir3/dir4/dir5/dir6/
/dir1/dir2/dir3/dir4/
I tried something like this:
awk -F'/*./.*$' '{print $0}' deep.list
but that didn't work out.
Your idea is smart, but needs a few corrections. Here's how you probably meant it:
awk -F'[^/]*/[^/]*$' '{print $1}' deep.list
Explanation:
First thing, you probably misspelled .* as *..
Then, the * modifier is greedy, therefore you need to take care it doesn't match more than you intended! The solution is simple, albeit slightly less readable: use [^/]* instead of .*. This way you match any characters except /.
Lastly, $0 represents the entire line, which hasn't been altered in the slightest by specifying a specially crafted custom field separator. In this case you want to print the first field instead: $1.
Here's a different approach than the two answers linked by @ender.qa:
awk '{gsub("[^/]+/[^/]+$","");print}' deep.list
And one employing a loop:
awk -F/ '{for(i=1;i<=NF-2;i++){printf "%s/",$i}; print ""}' deep.list
The substitution method is more easily implemented in Perl:
perl -lape 's"[^/]+/[^/]+$""' deep.list
or sed:
sed -E 's"[^/]+/[^/]+$""' deep.list
