Suppose I call a script with 3 arguments, a, abc, and xyz such that $@ contains these three arguments.
Suppose I want to call write a command:
command fooa fooabc fooxyz bara barabc barxyz
How would I accomplish that?
I don't think {foo,bar}$@ or {foo,bar}{$@} work since brace expansion happens before $@ is expanded.
您可以使用：
command "${@/#/foo}" "${@/#/bar}"
这使用了shell 参数扩展的替代变体。将#匹配锚定到参数的开头。
$ set -- a abc xyz
$ echo command "${@/#/foo}" "${@/#/bar}"
command fooa fooabc fooxyz bara barabc barxyz
$
有没有办法将其扩展到更大
{foo,bar,baz,...}$@或{foo,bar}{123,456}$@？
是的，但你最终会使用数组：
$ set -- a abc xyz
$ args=( "$@" )
$ for prefix in foo bar baz who why; do prefixed1+=( "${args[@]/#/$prefix}" ); done
$ echo command "${prefixed1[@]}"
command fooa fooabc fooxyz bara barabc barxyz baza bazabc bazxyz whoa whoabc whoxyz whya whyabc whyxyz
$
或者（注意你必须得到正确的排序）：
$ for prefix in 123 456; do prefixed1+=( "${args[@]/#/$prefix}" ); done
$ for prefix in foo bar; do prefixed2+=( "${prefixed1[@]/#/$prefix}" ); done
$ echo command "${prefixed2[@]}"
command foo123a foo123abc foo123xyz foo456a foo456abc foo456xyz bar123a bar123abc bar123xyz bar456a bar456abc bar456xyz
$
请注意，这会保留参数中的空格：
$ unset prefixed1 prefixed2
$ set -- 'a b' 'x y z'
$ args=( "$@" )
$ for prefix in 'p p-' '123 456 '; do prefixed1+=( "${args[@]/#/$prefix}" ); done
$ echo command "${prefixed1[@]}"
command p p-a b p p-x y z 123 456 a b 123 456 x y z
$ for prefix in 'foo 1-' 'bar 2-'; do prefixed2+=( "${prefixed1[@]/#/$prefix}" ); done
$ echo command "${prefixed2[@]}"
command foo 1-p p-a b foo 1-p p-x y z foo 1-123 456 a b foo 1-123 456 x y z bar 2-p p-a b bar 2-p p-x y z bar 2-123 456 a b bar 2-123 456 x y z
$ printf "[%s]\n" "${prefixed2[@]}"
[foo 1-p p-a b]
[foo 1-p p-x y z]
[foo 1-123 456 a b]
[foo 1-123 456 x y z]
[bar 2-p p-a b]
[bar 2-p p-x y z]
[bar 2-123 456 a b]
[bar 2-123 456 x y z]
$
第一个循环可能是：
for prefix in 'p p-' '123 456 '; do prefixed1+=( "${@/#/$prefix}" ); done
不使用${args[@]}数组。