在括号扩展范围内使用变量，将其馈送到for循环 [英] Using a variable in brace expansion range fed to a for loop

问题描述

这是myscript.sh

Here is myscript.sh

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

如果我运行myscript.sh 3，则输出为

3 {1..3}

代替

3 1
3 2
3 3

显然$3包含正确的值，所以for i in {1..$1}的行为为何与我直接编写for i in {1..3}的行为不同?

Clearly $3 contains the right value, so why doesn't for i in {1..$1} behave the same as if I had written for i in {1..3} directly?

推荐答案

您应该使用C样式的for循环来完成此操作:

You should use a C-style for loop to accomplish this:

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.

This avoids external commands and nasty eval statements.

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