如何将[！$number =~ ^[0-9]+$ ]改为BASH中的word选项？

Question

我的脚本中有一个输入，它写入运行某些软件所需的文件，但是该软件只允许两个可能的选项之一，即LO和LDO。

我已经编写的输入位于以下代码的顶部，底部的代码是建议我要操作的代码。

我想问“重叠(LO或LDO)：”和一个错误消息，如果其中一个选项没有输入并重复该步骤，而不是再次遍历整个脚本，而不是搜索0到9之间的值。

from我可以收集到什么，我需要更改^以包含LO和LDO，但是我不知道标点符号是什么，也不知道如何做到

### My Script ###
read -p "Overlap (LO or LDO): " OVERLAP

### Recommended Script ###
#!bin/bash
number=""
while [[ ! $number =~ ^[0-9]+$ ]]; do
    echo Please enter your age
    read number
done
echo You are $number years old

Answer 1

输入和验证循环通常如下所示：

while true; do
    read -p 'Overlap (LO or LDO): ' overlap

    case $overlap in
        LO|LDO) # ok
            break
            ;;
        *) # not ok
            echo 'Please enter LO or LDO' >&2
    esac
done

也就是说，当你验证答案是正确的时，你就有了一个无限循环。

很明显，你也可以这样做

while true; do
    read -p 'Overlap (LO or LDO): ' overlap

    if [ "$overlap" = LO ] || [ "$overlap" = LDO ]; then
        # ok
        break
    fi
    # not ok
    echo 'Please enter LO or LDO' >&2
done

或者，对于具有select的shell：

echo 'Select overlap:' >&2
select overlap in LO LDO; do
    case $REPLY in
        [12]) # ok
            break
            ;;
        *) # not ok
            echo 'Please select LO or LDO' >&2
    esac
done

Answer 2

您可以使用&&和操作符：

#! /bin/bash
overlap=""
first=1
while [[ $overlap != LO && $overlap != LDO ]] ; do
    if [[ ! $first ]] ; then
        echo 'Wrong answer, please answer LO or LDO!' >&2
    fi
    read -p 'Overlap (LO or LDO): ' overlap
    unset first
done

您还可以使用regex：

until [[ $overlap =~ ^LD?O$ ]] ; do
    ...

• ^与字符串的开头匹配。
• ?使前面的内容可选
• $与字符串的末尾匹配。

Answer 3

另一种方法

while read -p "Overlap (LO or LDO): " overlap;
do
 [[ "$overlap" =~ ^LD?O$ ]] && break || echo "Not LO or LDO"
done