while 变量不等于 x 或 y bash
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while variable is not equal to x or y bash
提问by Zvi
I'm trying to get user input. The input should be "1" or "2". for some reason I keep getting prompt even when I type 1 or 2.
我正在尝试获取用户输入。输入应为“1”或“2”。出于某种原因，即使我输入 1 或 2，我也会不断收到提示。
read -p "Your choice: " UserChoice
while [[ "$UserChoice" != "1" || "2" ]]
do
echo -e "\nInvalid choice please choose 1 or 2\n"
read -p "Your choice: " UserChoice
done
I'll appreciate your help Thanks!
我会很感激你的帮助 谢谢！
回答by chepner
!=does not distribute over ||, which joins two completeexpressions. Once that is fixed, you'll need to use &&instead of ||as well.
!=不分布 over ||，它连接了两个完整的表达式。一旦修复，您将需要使用&&而不是||。
while [[ "$UserChoice" != "1" && "$UserChoice" != "2" ]]
Actually, bashdoes support pattern matching which can be used similarly to what you had in mind.
实际上，bash确实支持模式匹配，可以类似于您想到的那样使用。
while [[ $UserChoice != [12] ]]
With the extgloboption set (which is on by default inside [[ ... ]]starting in bash 4.2, I believe), you can use something very close to what you originally had:
通过extglob选项设置（这是在默认情况下里面[[ ... ]]在bash 4.2开始，我相信），你可以使用的东西非常接近，你本来有哪些：
while [[ $UserChoice != @(1|2) ]]
