使用 if-else 在 Bash 中进行整数比较
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Integer comparison in Bash using if-else
提问by rahuL
I have a variable called choice. Now, I try to use if to compare the entered value:
我有一个名为choice. 现在，我尝试使用 if 来比较输入的值：
read $choice
if [ "$choice" == 2 ];then
#do something
elif [ "$choice" == 1 ];then
#do something else
else
echo "Invalid choice!!"
fi
The output goes directly to invalid choice if I enter either 1 or 2. I tried to put quotes around 1 and 2 inside the if statement. Still didn't work. Using -eqgives me an error "Unary operator expected".What am I doing wrong here?
如果我输入 1 或 2，输出将直接进入无效选择。我试图在 if 语句中将 1 和 2 周围的引号括起来。还是没用。使用-eq给我一个错误“一元运算符预期”。我在这里做错了什么？
回答by Mat
Your readline is incorrect. Change it to:
你的read线路不正确。将其更改为：
read choice
i.e. use the name of the variable you want to set, not its value.
即使用您要设置的变量的名称，而不是它的值。
-eqis the correct test to compare integers. See man testfor the descriptions (or man bash).
-eq是比较整数的正确测试。有关man test说明（或man bash），请参阅。
An alternative would be using arithmetic evaluation instead (but you still need the correct readstatement):
另一种方法是使用算术评估（但您仍然需要正确的read语句）：
read choice
if (( $choice == 2 )) ; then
echo 2
elif (( $choice == 1 )) ; then
echo 1
else
echo "Invalid choice!!"
fi
回答by devnull
For your example, caseseems to be what you probably want:
对于您的示例，case似乎是您可能想要的：
read choice
case "$choice" in
"1")
echo choice 1;
;;
"2")
echo choice 2;
;;
*)
echo "invalid choice"
;;
esac
回答by chepner
An unpractical answer, just to point out that since readtakes a variable name, the name can be specified by another variable.
一个不切实际的答案，只是为了指出，由于read采用了一个变量名，该名称可以由另一个变量指定。
choice=choice # A variable whose value is the string "choice"
read $choice # $choice expands to "choice", so read sets the value of that variable
if [[ $choice -eq 1 ]]; then # or (( choice == 1 ))
