bash if语句中的两个条件 [英] Two conditions in a bash if statement

问题描述

我正在尝试编写一个脚本，该脚本将读取两个选项，如果两个选项均为 y，我希望它说 Test Done！。如果它们中的一个或两个都不是 y，我想说测试失败！

I'm trying to write a script which will read two choices, and if both of them are "y" I want it to say "Test Done!" and if one or both of them isn't "y" I want it to say "Test Failed!"

这是我想出的：

echo "- Do You want to make a choice?"
read choice

echo "- Do You want to make a choice1?"
read choice1

if [ "$choice" != 'y' ] && [ "$choice1" != 'y' ]; then
    echo "Test Done!"
else
    echo "Test Failed!"
fi

但是当我用 y回答这两个问题时，它的意思是测试失败！ 而不是测试完成！。当我用 n回答两个问题时，都说测试完成！

But when I answer both questions with "y" it's saying "Test Failed!" instead of "Test Done!". And when I answer both questions with "n" it's saying "Test Done!"

我做错了什么？

推荐答案

您正在检查错误的条件。

You are checking for the wrong condition.

if [ "$choice" != 'y' ] && [ "$choice1" != 'y' ];

当 choice！='y'和 choice1！='y'，因此程序可以正确打印测试完成！ 。

The above statement is true when choice!='y' and choice1!='y', and so the program correctly prints "Test Done!".

更正后的脚本是

echo "- Do You want to make a choice ?"
read choice

echo "- Do You want to make a choice1 ?"
read choice1

if [ "$choice" == 'y' ] && [ "$choice1" == 'y' ]; then
    echo "Test Done !"
else
    echo "Test Failed !"
fi

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