bash 语法错误：预期的操作数（错误标记是“>”
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syntax error: operand expected (error token is "> "
提问by luke davis
./ex6.bash: ? 10: ((: > : syntax error: operand expected (error token is "> ")
And this is my code:
这是我的代码：
#!/bin/bash
printf "Input first number => "
read num1
printf "Input second number => "
read num2
num1=$a1
num2=$a2
if (( $a1>$a2 ))
then
while [ $a1==$a2 ];
do
let "a1 = $a1 - 1"
let "a2 = $a2 + 1"
if (( $a1==$a2 ))
then
printf " $num2 ~ $num1 mid point : $a1 \n"
break
elif (( $((a1 -1))==$a2 ))
then
printf " $num2 ~ $num1 mid point : $a1 \n"
break
fi
done
else
while [ $a1==$a2 ];
do
let "a1 = $a1 + 1"
let "a2 = $a2 - 1"
if (( $a1==$a2 ))
then
printf " $num1 ~ $num2 mid point : $a1 \n"
break
elif (( $((a1 -1))==$a2 ))
then
printf " $num1 ~ $num2 mid point : $a1 \n"
break
fi
done
fi
What's wrong and how do I fix it? I don't know what to do.
出了什么问题，我该如何解决？我不知道该怎么办。
回答by chepner
You never set a value for a1, so the arithmetic statement (($a1>$a2))expands to ((>)). Perhaps you meant a1=$num1instead of num1=$a1, but you don't need a1at all; you can just use $num1. The same holds for a2and num2.
您从未为 设置值a1，因此算术语句(($a1>$a2))扩展为((>))。也许您的意思是a1=$num1代替num1=$a1，但您根本不需要a1；你可以使用$num1. 这同样适用于a2和num2。
