bash - 在大括号扩展范围中使用变量馈送到 for 循环

Question

这是 myscript.sh

#!/bin/bash
for i in {1..$1};
do
    echo $1 $i;
done

如果我运行myscript.sh 3输出是

3 {1..3}

代替

3 1
3 2
3 3

显然$3包含正确的值，那么为什么不像for i in {1..$1}我for i in {1..3}直接写的那样表现呢？

Answer 1

您应该使用 C 风格的 for 循环来完成此操作：

for ((i=1; i<=$1; i++)); do
   echo $i
done

这避免了外部命令和讨厌的 eval 语句。

Answer 2

因为大括号扩展发生在变量扩展之前。http://www.gnu.org/software/bash/manual/bashref.html#Brace-Expansion。

如果你想使用大括号，你可以像这样严峻：

for i in `eval echo {1..$1}`;
do
    echo $1 $i;
done

摘要：Bash 是邪恶的。

Answer 3

您可以使用seq命令：

for i in `seq 1 $1`

或者你可以使用 C 风格for...loop：

for((i=1;i<=$1;i++))

Answer 4

Here is a way to expand variables inside braces without eval:

end=3
declare -a 'range=({'"1..$end"'})'

We now have a nice array of numbers:

for i in ${range[@]};do echo $i;done
1
2
3

Answer 5

I know you've mentioned bash in the heading, but I would add that 'for i in {$1..$2}' works as intended in zsh. If your system has zsh installed, you can just change your shebang to zsh.

Using zsh with the example 'for i in {$1..$2}' also has the added benefit that $1 can be less than $2 and it still works, something that would require quite a bit of messing about if you wanted that kind of flexibility with a C-style for loop.