Per man definition, this command gets the input from a file.
$ command -r FILENAME
Suppose that FILENAME is a file containing a list of filenames, as it was generated using ls > FILENAME.
How can I, instead, feed the command with the result of ls directly? In my head something like this should be possible:
$ ls | command -r
But it doesn’t, the output of ls doesn’t get hooked as an argument. Output:
Usage: command -r FILENAME error: -r option requires an argument
How could I obtain the desired effect?
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
This is dependent on the command. Some commands that read from a file expect that the file be a regular file, whose size is known in advance, and which can be read from any position and rewinded. This is unlikely if the contents of the file is a list of file names: then the command will probably be content with a pipe which it will just read sequentially from start to finish. There are several ways to feed data via a pipe to a command that expects a file name.
-
Many commands treat
-as a special name, meaning to read from standard input rather than opening a file. This is a convention, not an obligation.ls | command -r -
-
Many unix variants provide special files in
/devthat designate the standard descriptors. If/dev/stdinexists, opening it and reading from it is equivalent to reading from standard input; likewise/dev/fd/0if it exists.ls | command -r /dev/stdin ls | command -r /dev/fd/0
-
If your shell is ksh, bash or zsh, you can make the shell deal with the business of allocating some file descriptor. The main advantage of this method is that it’s not tied to standard input, so you can use standard input for something else, and you can use it more than once.
command -r <(ls)
-
If the command expects the name to have a particular form (typically a particular extension), you can try to fool it with a symbolic link.
ln -s /dev/fd/0 list.foo ls | command -r list.foo
Or you can use a named pipe.
mkfifo list.foo ls >list.foo & command -r list.foo
Note that generating a list of files with ls is problematic because ls tends to mangle file names when they contain unprintable characters. printf '%sn' * is more reliable — it’ll print every byte literally in file names. File names containing newlines will still cause trouble, but that’s unavoidable if the command expects a list of file names separated by newlines.
Method 2
It should be:
ls | xargs -n 1 command -r
Edit: for names with blank spaces:
ls | xargs -d 'n' -n 1 command -r
Method 3
Many commands accept - as a “filename” that means “use the standard input”, but this convention is far from universal. Read the man page.
Method 4
Actually, the only reliable solution that I know of that can handle all filenames, including those with a newline in them, is:
find . -maxdepth 1 -print0 | xargs -n 1 -0 command -r
The only non allowed character in the filename in this case is the null character, which isn’t allowed in filenames anyway.
Method 5
This should work for your purpose: ls | command -r /dev/stdin
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0