piping output of list to cp

andy.l · 05-16-2011, 04:58 PM

Hi
I have the following code

Code:

ls ~/ | grep ^[[:lower:]]

and would like to take the result of this output and use this as the basis for a cp command to move these files to another folder. How can I easily achieve this?

/Andy

Chirel · 05-16-2011, 05:08 PM

Even if it's ugly, and if you have no space in filenames you can do something like this

Code:

for i in $(ls ~/ | grep ^[[:lower:]]); do cp $i newdestdir; done

andy.l · 05-16-2011, 05:35 PM

But if there are spaces in filenames? Hos do i handle thore?

/Andy

David the H. · 05-16-2011, 06:02 PM

You should almost never need to use ls to get a list of files. Bash has excellent globbing ability, and can do almost everything you need. Particularly when you use extended globbing patterns.

Code:

cp [[:lower:]]* targetdirectory

Edit: Note also that word splitting isn't done on the results of a glob, at least as long as it's being passed directly to the program being executed (splitting happens before globbing), so spaces and other reserved characters shouldn't be a problem. This wouldn't be true if additional processing of the line is done after the globbing substitution is made, such as when using it inside a $() command substitution.