Shell Programming and Scripting

Another way of doing this would be:

# Assume file with paths to files are in file /tmp/somefiles
for filename in $(cat /tmp/somefiles) ; do
mkdir -p $(dirname $filename)
done

This should work in ksh and bash.
/fimblo

Franklin52 can u pls explain the sed command. I have never used this command. I want to learn abt it . Pls help me to understand this.

I got from man pages that it is use to replace strings. But how it is working here.

Thanks in Advance
Joy

It replaces the file name with the part of the file name which comes before the last slash, i.e. the directory part of the name.

Hi Joy,

Code:

 sed 's!\(.*/\).*!mkdir -p \1!' file | sh

Here in the mentioned code Franklin52 has used ! as a delimiter ( may be thats confusing you).

And the sed body goes like this.

Sed "DELIM" pattern "DELIM" Command "DELIM" input

sed s "DELIM" \(.*/\).* "DELIM" mkdir -p \1 "DELIM" file


And the \(.*/\).* part matches strings between "/"a-z"/" n no of times.

Hope you understood that,

Keep Smiling,
Harsha.

Thanks Harsha for helping me. But I have one doubt here, how \(.*/\).* is equal to "/"a-z"/".

In sed you can save portions of a string with \(.*\) and recall them back with \1, \2, \3 etc.
Here within \( and \) we save a portion from the start .* of a line until the last /.

Regards