Linux bash:多变量赋值 [英] Linux bash: Multiple variable assignment

问题描述

Linux bash 中确实存在类似于 PHP 中的以下代码的内容:

Does exist in linux bash something similar to the following code in PHP:

list($var1, $var2, $var3) = function_that_returns_a_three_element_array() ;

即你在一个句子中为 3 个不同的变量分配了一个对应的值.

i.e. you assign in one sentence a corresponding value to 3 different variables.

假设我有一个 bash 函数 myBashFuntion 将字符串qwert asdfg zxcvb"写入标准输出.是否可以执行以下操作:

Let's say I have the bash function myBashFuntion that writes to stdout the string "qwert asdfg zxcvb". Is it possible to do something like:

(var1 var2 var3) = ( `myBashFuntion param1 param2` )

等号左边的部分当然不是有效的语法.我只是想解释一下我的要求.

The part at the left of the equal sign is not valid syntax of course. I'm just trying to explain what I'm asking for.

不过，以下是有效的:

array = ( `myBashFuntion param1 param2` )
echo ${array[0]} ${array[1]} ${array[2]}

但是索引数组不像普通变量名那样具有描述性.
但是，我可以这样做:

But an indexed array is not as descriptive as plain variable names.
However, I could just do:

var1 = ${array[0]} ; var2 = ${array[1]} ; var3 = ${array[2]}

但这是我更希望避免的另外 3 个陈述.

But those are 3 more statements that I'd prefer to avoid.

我只是在寻找一种快捷语法.可能吗?

I'm just looking for a shortcut syntax. Is it possible?

推荐答案

我想到的第一件事:

read -r a b c <<<$(echo 1 2 3) ; echo "$a|$b|$c"

输出是，不出所料

1|2|3

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