问题
I have a script P which accepts file names as parameters:
P file1 file2 file3 ....
I also have a script G which generates a (typically short list) of file names, one file name per line. In a master script which I would like to write in zsh, I want to use G to generate the file names to be processed by P. The naive attempt goes like this:
P $(G)
This works nearly well, only that I'm living in a world where malicious people enjoy creating files with embedded spaces. If G would generate the list of files like this:
one_file
a file with spaces
P would be called as
P one_file a file with spaces
instead of
P 'one_file' 'a file with spaces'
One obvious solution would be to glue G and P together not by command substitution, but by a small program in some language (Ruby, Perl, Python, Algol68,....), which does an exec of P and passes the parameters to a read from stdin.
This would be trivial to write and could even be made reusable. However, I wonder, whether zsh with its grabbag of goodies would not have a solution for this problem builtin?
回答1:
This can be achieved with:
P ${(f)"$(G)"}
which will call P as
P 'one_file' 'a file with spaces'
if the output of G is
one_file
a file with spaces
Explanation:
The double quotes around $(G) tell zsh to take the output of G as one word (in the shell sense of the word word).
So just calling
P "$(G)"
would be equivaent of
P 'one_file
a file with spaces'
This is where the Parameter Expansion Flag f comes into play. ${(f)SOMEVAR} tells zsh to split $SOMEVAR into words at newlines. Instead of a parameter (like SOMEVAR) one can also use a command substitution (here $(G)) inside ${...} type parmeter expressions, which leads to the the command P ${(f)"$(G)"}.
来源：https://stackoverflow.com/questions/29028828/zsh-command-substitution-and-proper-quoting