在 Try 中的“if”语句中使用“或”，在 Python 中除外

Question

I want to make the loop continue and catch errors like entering "bob" as an input or integers less than 1 and greater than 4.

salesNum = input("How many sales persons to process? ")
for num in salesNum:
    salesName = input("Enter Salesperson Name: ")
    salesLevel = 0
    while salesLevel == 0:
        try:
            salesLevel = int(input("Enter Salesperson Level: "))
            if salesLevel < 1 or salesLevel > 4:
                print("error, try again.")
                salesLevel = int(input("Enter Salesperson Level: "))
        except ValueError:
            print("error, try again")
            continue
    hoursWorked = float(input("Enter Hours Worked: "))
    salesAmount = float(input("Enter Number of Sales: "))
print("Clear")

It works fine if I make one error, but multiple errors don't get caught and the program moves on to the next questions. I have tried breaking the if statement up, but that doesn't solve the issue.

Answer 1

The error won't be ValueError , so you either change it to TypeError :

except TypeError:

Or to any error:

except:

Answer 2

salesNum = input("How many sales persons to process? ")
for num in salesNum:
    salesName = input("Enter Salesperson Name: ")
    salesLevel = 0
    while salesLevel == 0:
        try:
            salesLevel = int(input("Enter Salesperson Level: "))
            if salesLevel < 1 or salesLevel > 4:
                print("error, try again.")
                salesLevel = int(input("Enter Salesperson Level: "))
        except Exception:
            print("error, try again")
            continue
    hoursWorked = float(input("Enter Hours Worked: "))
    salesAmount = float(input("Enter Number of Sales: "))
print("Clear")

try change ValueError to Exception this will catch every abnormality