堅持一個簡單的Python程序「ELIF」

Question

你好我有一個麻煩，在做我的簡單的計算器代碼：d

def cal(): 

while True: 

print ("welcome to my calculator!") 
print("choose an operation") 

op = input(" +, - ,/ ,*") 


if op == "+": 

    num1 = float(input("enter your first number:")) 
    num2 = float(input("enter your second number:")) 

    print(str(num1 + num2) 



elif op == "/": 

    num1 = float(input("enter your first number:")) 
    num2 = float(input("enter your second number:")) 

    print(str(num1/num2) 


else: 

break 



cal() 

當過我運行的代碼，它說無效語法在ELIF

這裏有什麼問題？

Answer 1

你錯過了一堆括號。如果你想利用你的程序還用這個作爲一個模型：

# This function adds two numbers 
def add(x, y): 
    return x + y 

# This function subtracts two numbers 
def subtract(x, y): 
    return x - y 

# This function multiplies two numbers 
def multiply(x, y): 
    return x * y 

# This function divides two numbers 
def divide(x, y): 
    return x/y 

print("Select operation.") 
print("1.Add") 
print("2.Subtract") 
print("3.Multiply") 
print("4.Divide") 

# Take input from the user 
choice = input("Enter choice(1/2/3/4):") 

num1 = int(input("Enter first number: ")) 
num2 = int(input("Enter second number: ")) 

if choice == '1': 
    print(num1,"+",num2,"=", add(num1,num2)) 

elif choice == '2': 
    print(num1,"-",num2,"=", subtract(num1,num2)) 

elif choice == '3': 
    print(num1,"*",num2,"=", multiply(num1,num2)) 

elif choice == '4': 
    print(num1,"/",num2,"=", divide(num1,num2)) 
else: 
    print("Invalid input") 

Answer 2

您從未關閉打印功能中的括號。其他if聲明也一樣。您以後也應該使用4個空格的縮進。

if op == "+": 
    num1 = float(input("enter your first number:")) 
    num2 = float(input("enter your second number:")) 
    print(str(num1 + num2))