Shell 破折号循环跳过前两个 arguments

Question

How to skip first two arguments using "$@" in shell dash and use the rest of arguments in for loop?

For example, I have something like command --option file1 file2 file3 . I want to skip first two arguments and loop over all the files (any number of files).

I have tried using the following line:

files=$(echo "$@" | awk '{for (i=2; i<=NF; i++) print $i}' | tr -d '\n')

However it does not provide a line of files, but something like file1file2file3 .

I would like to be able to iterate over each file name individually later in the for loop:

for file in "$files"
do
    rm "./desktop/temp/$file"
done

Answer 1

I assume that by "shell dash" you mean that you are using Dash , a light-weight shell that sticks to the Posix standard. Comparing to Bash, Dash is quite limited and doesn't offer functionalities like arrays or complex parameter substitutions like ${@:m:n} .

I also assume that, by doing files=$(echo "$@" |...) you know for sure that none of your arguments contains space characters.

That said, your issue is due to the | tr -d '\n' | tr -d '\n' part which removes the end of line that serves as a word separator. That's why you get "file1file2file3".

But the whole command can be enhanced. Here is a better Posix-compatible way of doing this:

files=$(shift 2; printf '%s ' "$@")

( shift being done in a sub-shell, it doesn't affect the argument list in the current shell)

You may even do this:

(
    shift 2
    for file in "$@"; do
        rm "./desktop/temp/$file"
    done
)

(here the whole loop is executed in a subshell)

... or write a function:

rm_files()
{
    shift 2
    for file in "$@"; do
        rm "./desktop/temp/$file"
    done
}

rm_file "$@"

One last suggestion: is it really necessary to keep arguments 1 and 2 in the argument list? You could do something like:

arg1="$1"
arg2="$2"
shift 2
for file in "$@"; do
    rm "./desktop/temp/$file"
done

... and if you need all your arguments at once, you can do:

for arg in "$arg1" "$arg2" "$@"; do ...; done