How to remove last part of the line with command?

[abcuser]

I got Windows file with file paths in it. I need to retain only directory structure and remove file names.

Sample data in myfile.txt:

C:\aaa\bbbb\cccc\xxx.txt
C:\aaa\bbbb\abc.txt
C:\aaa\zzzz.txt
<many hundreds of lines with different path depth>

So in myfile.txt I need to search for the last backslash character "\" and remove everything to the end of line, to get:

C:\aaa\bbbb\cccc
C:\aaa\bbbb
C:\aaa

How to remove file names from path?
Thanks

[Lars Noodén]

I tried basename and dirname, but they did not work. However, sed does the job.

sed -e 's/\\[^\\]*$//' < myfile.txt > output.txt

[evilsoup]

I think you'll have to use sed for this.

cat filename | sed 's\(.*\\\)/\1/' > outputfile

will give you an output file with the xxx.txt removed, but it keeps the final backslash, I don't know how to remove that.

EDIT: nevermind, ninja'd with a better solution

[steeldriver]

or you could use bash substitution

while read -r path; do echo "${path%\\*}"; done < myfile.txt