你如何防止“ValueError：int（）的无效文字，base 10：''，定义函数？

Question

我正在尝试猜数字游戏。我试图让用户得到 3 次猜测，如果猜测太大，代码会说它太大，如果猜测太小，那么代码应该告诉他们，同时也允许如果他们的答案超出 1-10 的范围，他们重新输入他们的猜测，因为那是随机生成的数字的来源。我计划添加其他级别和排行榜，所以如果有人可以帮助我，我将不胜感激。这是我的代码：

import time
import random

answer1= ""
answer2= ""
answer3= ""

one= "You have two more guesses: "
two= "You only have one more guess. Be careful: "

number= random.randint(1,10)
number=str(number)

def code(a,number,b,n,d,e):
    a = int(a)
    number= int(number)
    if a > number and a <11:
        b=input("the number is too big. "+ n )
    elif a< number and a>-1:
        d= input("the number is too big. "+ n )
    elif a> 10 or a<0:
        e=input("please input a number smaller than 10 and bigger than 0: ") 


print(" -------------------- Guess the number game --------------------")
time.sleep(1)
answer= input("The aim of the game is to guess the number. Are you ready? (yes/no): ").lower()

if "yes" in answer:
    answer1=input("I am thinking of a number, between 0 and 10, What is the number: ")

    if answer1 == number:
        print ("Correct! Wow.. it only took you one guess")
        time.sleep(1)
        print("Come back again sometime...")
        time.sleep(1)
        print("BYE!")
        time.sleep(1)
        quit()


    elif answer1 != number:
        code(answer1,number,answer2,one,answer2,answer2)

    if answer2 == number:
        print ("Correct! it took you two turns.")
        time.sleep(1)
        print("Come back again sometime")
        time.sleep(1)
        print("BYE!")
        time.sleep(1)
        quit()


    elif answer2 != number:
        code(answer2,number,answer3,two,answer3,answer3)


    if answer3 == number:
        print ("Correct! Phew.... you guessed correctly in your last guess!")
        time.sleep(1)
        print("Come back again sometime")
        time.sleep(1)
        print("BYE!")
        time.sleep(1)
        quit()

    elif answer3 != number:
        print("Incorrect..... Sorry! ")
        time.sleep(1)
        print("The correct number was "+number)
        time.sleep(1)
        print("You lose! But don't stop trying!")
        time.sleep(1)
        quit()


elif "yes" != answer:
    print("That's too bad...")
    time.sleep(1)
    print("Come back when you're ready to....")
    time.sleep(2)
    print("LOSE!!")
    time.sleep(1)
    quit()

当我像这样使用它时，它一直给我：“ValueError：int() 的无效文字，基数为 10:''，当我这样使用它时（这是在 shell 中）：

 ------------------- Guess the number game --------------------
The aim of the game is to guess the number. Are you ready? (yes/no): yes
I am thinking of a number, between 0 and 10, What is the number: 2
the number is too big. You have two more guesses: 1
Traceback (most recent call last):
  File "C:/Python33/Guweesss da numba.py", line 63, in <module>
    code(answer2,number,answer3,two,answer3,answer3)
  File "C:/Python33/Guweesss da numba.py", line 19, in code
    a = int(a)
ValueError: invalid literal for int() with base 10: ''

忽略代码的奇怪名称 - 我之前已经尝试过多次，所以没有使用任何名称。

Answer 1

你的问题是 python 没有像你想象的那样传递变量

def code(a,number,b,n,d,e):
    a = int(a)
    number= int(number)
    if a > number and a <11:
        b=input("the number is too big. "+ n )
    elif a< number and a>-1:
        d= input("the number is too big. "+ n )
    elif a> 10 or a<0:
        e=input("please input a number smaller than 10 and bigger than 0: ") 

无论您以b、d 和e 传递的内容在您执行e=input(...) 时都不会得到更新，因此当您完成时answer2 仍然是""。也没有理由将同一个变量传递 3 次。

你需要重做这个函数才能返回结果。

def code(a,number,n):
    a = int(a)
    number= int(number)
    if a > number and a <11:
        b=input("the number is too big. "+ n )
    elif a< number and a>-1:
        b= input("the number is too big. "+ n )
    elif a> 10 or a<0:
        b=input("please input a number smaller than 10 and bigger than 0: ") 
    return b

然后通过以下方式调用：

answer2 = code(answer1, number, one)

见repl.it

一些进一步的说明：

• 您可能希望按照其他人的建议添加输入错误检查
• 对于太大和太小的猜测，您都说“数字太大”
• 您可能需要查看一个循环，以便在不重新启动的情况下继续播放。
• 我会推荐一个关于函数和传递变量的教程，因为这是根本问题。

Answer 2

分析

您正在尝试将空字符串转换为整数。繁荣！ 如果您真的想要这样的功能，请将所有三个猜测初始化为“-1”之类的值，这样这些值在语法上是合法的，但在语义上超出了范围。

改进检查

if a.isdigit():
    a = int(a)
else:
    print ("Don't be silly!  I need a number!")

建议

在测试之前不要编写太多代码。写几行，测试它们，直到他们做你想做的事才继续。在这里，您试图在 85 行代码中找出一个小错误。

了解函数的工作原理。您的code 函数接受六个参数，但忽略b、d 和e，覆盖输入的任何值。我认为您可能将那些与“可变”值混淆了，您可以更改这些值。调用程序不会收到分配的值。

使用有意义的变量名。

Answer 3

你可以捕捉到异常：

try:
    a = int(a)
exception ValueError as error:
    print(error)
    print('"{}" is not an integer'.format(a))

Answer 4

您可能想尝试一种非常 Python 的方式来检查变量，方法是使用 try ... except 构造。所以改为 a = int(a) 是这样的：

try:
  a = int(a)
except:
  print ... something ...

好吧，正如@TemporalWolf 指出的那样，功能存在很大问题。所以看看吧。 最后，您尝试获取数字并检查它，因此您可能需要一个循环，直到用户提供所需范围内的数字。它看起来像这样

a=-1
while 0<a<10:
  answer=input("I am thinking of a number, between 0 and 10, What is the number: ")
  try:
    a = int(answer)
  except:
    print "I need a number!"
    a = -1
    continue
  else:
    if a<1:
      print "number is negative"
      continue
    if a>10:
      print "number too big"
      continue

这将保证您输入的内容正确，并且您有一个从 1 到 10 的数字。

注意，@Prune 解决方案也非常好（老实说，我会使用它：）