如何将所有命令参数放在一个变量中 [英] how to put all command arguments in one variable
问题描述
我想执行一个需要3个参数的shell脚本.
I want to execute a shell script that require 3 arguments.
参数2包含一个带空格的字符串
The argument number 2 contains a string with space
我想将所有参数都放在一个这样的变量中:
I want to put all arguments in one variable like this:
Linux:~# kk="\"111\" \"222 222\" \"333\""
Linux:~# echo $kk
"111" "222 222" "333"
现在,如果我调用一个函数:
Now If I call a function:
func() {
echo ---$1---
echo ---$2---
echo ---$3---
}
通过$ kk变量以这种方式
with the $kk variable in this way
func $kk
然后它将返回
Linux:~# func $kk
---"111"---
---"222---
---222"---
我期待得到这个结果
---111---
---222 222---
---333---
如何在不使用eval
的情况下解决此问题?
How to solve this issue without using eval
?
我知道eval
解决了这个问题,但是我不想使用它(因为如果我多次执行这样的调用会花费时间).
I know that the eval
solve this issue but I do not want to use it (since it takes time if I execute many time a such call).
推荐答案
如果功能(如数组)没有使bash表现得比POSIX sh更具表现力,则没有理由添加它们. :)
If features (like arrays) didn't make bash more expressive than POSIX sh, there would have been no reason to add them. :)
也就是说,您可以通过根据需要覆盖"$@"
来解决此问题:
That said, you can work around it by overriding "$@"
for your needs:
set -- 111 "222 222" 333
printf '%s\n' "$@"
...将打印...
...will print...
111
222 222
333
如果您需要逐步构建参数列表,则可以使用"$@"
作为第一个参数来多次调用set
(确保您遵守引号!):
If you need to build the arguments list step by step, you can make several calls to set
with "$@"
as the first arguments (make sure you observe the quotes!):
set -- 111
set -- "$@" "222 222"
set -- "$@" 333
printf '%s\n' "$@"
...将打印...
...will print...
111
222 222
333
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