Why does echo * | rm -f not work? as per my understanding, echo * should list all the files in the directory and pipe should send this list as an argument to the rm -f command, which should delete the files. What am I missing?
2 Answers
Pipe sends data to stdin. And not to arguments list. These are two very different concepts.
Arguments are what is in line after command line. Stdin is basically just like a file that is already opened, from which you can read.
Please also note that doing things like this: echo * | xargs rm -f is bad idea because it might do weird things in case of files with spaces in names, or with "-" at the beginning.
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Thank you very much. I'm going to explore all the points you touched.– LavyaJun 21, 2013 at 11:20
Pass output of echo * as arguments rather than stdin inputs from pipe.
rm -f `echo *`
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4This is very, very bad idea. Consider file named "-r ." Of course it would solve the problem the poster is asking, but doing this is just plain wrong.– user7385Jun 21, 2013 at 11:20
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@depesz then
rm -f -- $(echo *)will do the trick (if you have a smart enoughrm). Otherwiserm -f $(echo ./*)will do. Jun 21, 2013 at 11:39 -
@depesz add
--to solve this.rm -f -- $(echo *)– Arie ShawJun 21, 2013 at 11:40 -
3This solves parameter problem, but still can yield bad results in case of files with spaces in names. Really - there is no reason to use echo, and it will cause pain in any way.– user7385Jun 21, 2013 at 11:54
echo *doesn't list hidden files butecho * .*does.echo .*you'll also have.and..which isn't a good idea at all. Withshopt -s dotglob,echo *will expand to all files, including the hidden ones, but not.nor..; it will be safer.