bash 将传递的参数存储在单独的变量中 -shell 脚本
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storing passed arguments in separate variables -shell scripting
提问by Nathan Pk
In my script "script.sh" , I want to store 1st and 2nd argument to some variable and then store the rest to another separate variable. What command I must use to implement this task? Note that the number of arguments that is passed to a script will vary.
在我的脚本 "script.sh" 中，我想将第一个和第二个参数存储到某个变量，然后将其余的存储到另一个单独的变量。我必须使用什么命令来执行此任务？请注意，传递给脚本的参数数量会有所不同。
When I run the command in console
当我在控制台中运行命令时
./script.sh abc def ghi jkl mn o p qrs xxx #It can have any number of arguments
In this case, I want my script to store "abc" and "def" in one variable. "ghi jkl mn o p qrs xxx" should be stored in another variable.
在这种情况下，我希望我的脚本将“abc”和“def”存储在一个变量中。“ghi jkl mn op qrs xxx”应该存储在另一个变量中。
回答by William Pursell
If you just want to concatenate the arguments:
如果您只想连接参数：
#!/bin/sh
first_two=" " # Store the first two arguments
shift # Discard the first argument
shift # Discard the 2nd argument
remainder="$*" # Store the remaining arguments
Note that this destroys the original positional arguments, and they cannot be reliably reconstructed. A little more work is required if that is desired:
请注意，这会破坏原始位置参数，并且无法可靠地重建它们。如果需要，还需要做更多的工作：
#!/bin/sh
first_two=" " # Store the first two arguments
a=""; b="" # Store the first two argument separately
shift # Discard the first argument
shift # Discard the 2nd argument
remainder="$*" # Store the remaining arguments
set "$a" "$b" "$@" # Restore the positional arguments
回答by Ignacio Vazquez-Abrams
Slice the $@array.
切片$@数组。
var1=("${@:1:2}")
var2=("${@:3}")
回答by Iluvatar
Store all args in table
将所有参数存储在表中
vars=("$@")
echo "${vars[10]}"
