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I am trying to run a program like this:
$CMD $ARGS
where $ARGS is a set of arguments with spaces. However, zsh appears to be handing off the contents of $ARGS as a single argument to the executable. Here is a specific example:
$export ARGS="localhost -p22"
$ssh $ARGS
ssh: Could not resolve hostname localhost -p22: Name or service not known
Is there a bash or zsh flag that controls this behavior?
Note that when I put this type of command in a $!/bin/sh script, it performs as expected.
Thanks,
SetJmp
883
You can pass more than one argument to your bash script. In general, here is the syntax of passing multiple arguments to any bash script: script.sh arg1 arg2 arg3 … The second argument will be referenced by the $2 variable, the third argument is referenced by $3 , .. etc.
Assigning the arguments to a regular variable (as in args="$@" ) mashes all the arguments together like "$*" does. If you want to store the arguments in a variable, use an array with args=("$@") (the parentheses make it an array), and then reference them as e.g. "${args[0]}" etc.
You can handle command-line arguments in a bash script in two ways. One is by using argument variables, and another is by using the getopts function.
Bash users need to understand that alias cannot take arguments and parameters. But we can use functions to take arguments and parameters while using alias commands.
It will work if you use eval $CMD $ARGS.
183
In zsh it's easy:
Use cmd ${=ARGS} to split $ARGS into separate arguments by word (split on whitespace).
Use cmd ${(f)ARGS} to split $ARGS into separate arguments by line (split on newlines but not spaces/tabs).
As others have mentioned, setting SH_WORD_SPLIT makes word splitting happen by default, but before you do that in zsh, cf. http://zsh.sourceforge.net/FAQ/zshfaq03.html for an explanation as to why whitespace splitting is not enabled by default.
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