当说类型错误时，我该如何解决它：+：'type'和"类型"不支持的操作数类型？

Question

我知道这有问题，类型有问题，但我忘记了如何更改它。

number = int(input("Which number do you want to check? "))
if number/2 == int + float:
print("This number is an odd number")
else:
print("This number is an even number")

Answer 1

不能将数据类型用作操作数。请改用任何变量。让我为你做这个

number = int(input("Which number do you want to check? "))
if number % 2 != 0:
print("This number is an odd number")
else:
print("This number is an even number")

Answer 2

正如错误所说的"不能对"type"one_answers"types"操作+，不能同时使用int+float，因为实际上这没有意义。当你进行比较时，你需要将其与固定编号进行比较。int不定义任何固定编号。尝试将提醒与零进行比较，因为任何偶数除以2都不会给出提醒。

类似这样的东西：

if number%2 == 0:
print("This number is an even number")
else:
print("This number is an odd number")

Answer 3

语句类型&试图比较的类型是如果integer+float等于var(number(/2本质上是false，那么在比较时应该执行var1+var2。

答案：

number = int(input("Which number do you want to check? "))
if (number % 2) > 0:
print("This number is an odd number")
else:
print("This number is an even number")

如何接近等于示例：

number = int(input("Which number do you want to check? "))
if number == 1+2:
print("This is correct")
else:
pass

Answer 4

我修复了它，现在它变为

number = int(input("Which number do you want to check? "))
if number / 2 == 0:
print("This number is an even number")
else:
print("This number is an odd number")