BASH中select菜单结合if语句失效求助：仅输出周一内容

你遇到的问题核心是Bash中test命令（[）的比较语法错误：

在[的判断逻辑里，==两边必须用空格分隔，否则整个$day==monday会被当成一个完整的非空字符串。而test命令对非空字符串的判定永远为真，所以第一个if分支会一直触发，不管你实际选择了什么选项。

修正后的脚本 #

把每个判断条件里的==前后加上空格，同时给变量和匹配字符串加上双引号（这是Bash脚本的好习惯，能避免变量为空时出现语法错误）：

select day in monday tuesday wednesday thursday friday saturday sunday;
do
    echo "当前选择：$day"
    if [ "$day" == "monday" ]; then
        echo "need a lot of coffee"
    elif [ "$day" == "tuesday" ]; then
        echo "need even more coffee"
    elif [ "$day" == "wednesday" ]; then
        echo "need something stronger"
    elif [ "$day" == "thursday" ]; then
        echo "need tequila"
    elif [ "$day" == "friday" ]; then
        echo "regret thursday"
    elif [ "$day" == "saturday" ]; then
        echo "go for a run"
    else
        echo "try not to think about tomorrow"
    fi
done

为什么case语句没问题？ #

case语句的语法是基于模式匹配，它的判断逻辑和test命令完全不同，不需要严格的空格分隔要求，所以你用case实现的版本能正常工作。

内容的提问来源于stack exchange，提问作者Satopu