构建包含空格的参数列表 [英] build argument lists containing whitespace

问题描述

在 bash 中，可以转义包含空格的参数.

In bash one can escape arguments that contain whitespace.

foo "a string"

这也适用于命令或函数的参数:

This also works for arguments to a command or function:

bar() {
    foo "$@"
}

bar "a string"

到目前为止一切顺利，但如果我想在调用 foo 之前操作参数怎么办?

So far so good, but what if I want to manipulate the arguments before calling foo?

这不起作用:

bar() {
    for arg in "$@"
    do
        args="$args \"prefix $arg\""
    done

    # Everything looks good ...
    echo $args

    # ... but it isn't.
    foo $args

    # foo "$args" would just be silly
}

bar a b c

那么当参数包含空格时如何构建参数列表?

So how do you build argument lists when the arguments contain whitespace?

推荐答案

(至少)有两种方法可以做到这一点:

There are (at least) two ways to do this:

1. 使用数组并使用"${array[@]}"扩展它:

bar() {
    local i=0 args=()
    for arg in "$@"
    do
        args[$i]="prefix $arg"
        ((++i))
    done

    foo "${args[@]}"
}

那么，我们学到了什么?"${array[@]}" 对 ${array[*]} 什么是 "$@" 对 $*.

So, what have we learned? "${array[@]}" is to ${array[*]} what "$@" is to $*.

或者如果你不想使用数组，你需要使用eval:

Or if you do not want to use arrays you need to use eval:

bar() {
    local args=()
    for arg in "$@"
    do
        args="$args \"prefix $arg\""
    done

    eval foo $args
}

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