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How to create a directory structure with getting input from a file.

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Top Forums Shell Programming and Scripting How to create a directory structure with getting input from a file.
# 8  
Old 09-23-2008
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)

This should work in ksh and bash.
# 9  
Old 09-29-2008

Originally Posted by Franklin52
Assuming you want to create the directories:


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


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
# 10  
Old 09-29-2008
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.
# 11  
Old 09-29-2008
Hi Joy,

 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,
# 12  
Old 09-29-2008
Originally Posted by harshakirans

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

Thanks Harsha for helping me. But I have one doubt here, how \(.*/\).* is equal to "/"a-z"/". Smilie
# 13  
Old 09-29-2008
Originally Posted by itsjoy2u
Thanks Harsha for helping me. But I have one doubt here, how \(.*/\).* is equal to "/"a-z"/". Smilie
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 /.


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