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I am wondering how arguments given to a function in bash can be properly "forwarded" to another function or program.
For example, in Mac OS X there is a command line program open (man page) that will open the specified file with its default application (i.e. it would open a *.h file in Xcode, or a folder in Finder, etc). I would like to simply call open with no arguments to open the current working directory in Finder, or provide it the typical arguments to use it normally.
I thought, "I'll just use a function!" Hah, not so fast there, I suppose. Here is what I've got:
function open
{
if [ $# -eq 0 ]; then
/usr/bin/open .
else
/usr/bin/open "$*"
fi
}
Simply calling open works great, it opens the working directory in Finder. Calling open myheader.h works great, it opens "myheader.h" in Xcode.
However, calling open -a /Applications/TextMate.app myheader.h to try to open the file in TextMate instead of Xcode results in the error "Unable to find application named ' /Applications/TextMate.app myheader.h'". It seems passing "$*" to /usr/bin/open is causing my entire argument list to be forwarded as just one argument instead.
Changing the function to just use usr/bin/open $* (no quoting) causes problems in paths with spaces. Calling open other\ header.h then results in the error "The files /Users/inspector-g/other and /Users/inspector-g/header.h do not exist", but solves the other problem.
There must be some convention for forwarding arguments that I'm just missing out on.
261
Arguments can be passed to the script when it is executed, by writing them as a space-delimited list following the script file name. Inside the script, the $1 variable references the first argument in the command line, $2 the second argument and so forth. The variable $0 references to the current script.
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
Example of command substitution using $() in Linux: Again, $() is a command substitution which means that it “reassigns the output of a command or even multiple commands; it literally plugs the command output into another context” (Source).
You indeed missed "$@", which is designed for this case.
196
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