How do you write a bash script function that is dynamic to the parameters passed?

Question

I have a bash function (slightly simplified for explaining)

copy_to() {
    cp $1 $2 $3
}

This works fine:

copy_to -f /demo/example1.xml new_demo/

But Let's say I want to copy all the xml files, the following code will have issues:

copy_to -f /demo/*.xml new_demo/

Obviously I could just write cp -f /demo/*.xml new_demo/, but is there anyway to get the copy_to function to work for a list of files (which passes more than just 3 parameters) as well as a single file?

Answer 1

There are $@ and $* which contain a list of all parameters. You should use $@, because it works inside of double quotes. Else, file names containing spaces would break your code.

copy_to() {
    cp "$@"
}

If one of the parameters is special, you can use the shift command to remove it from the list of parameters, like so:

example() {
    destination="$1"
    shift
    echo "copying $@ to $destination"
}

shift removes the first parameter from the list, therefore you’ll have to save it in another location first. After calling shift, the contents of $1 will be what was $2, $2 will contain what was $3 and so on. $@ expands to all parameters (excluding those that were removed by shift).

Note that you cannot shift parameters off the end of your parameter list, only from the beginning.

Answer 2

As Scytale said $0 and $* contain a list of all parameters. $# contains param count.

You can consider use getopts command for parameter parsing.

Best regards

Answer 3

If you need to iterate over the arguments:

f() {
    for arg
    do
        do_something $arg
    done
}

The in $@ is implied in for arg (explicitly: for arg in $@).