Eclipse 为什么程序在执行前多次询问我输入？

我一直在做一个练习，让我对基础知识更加熟悉，但这个程序不会像我希望或期望的那样运行。 我已经使用输入等编写了类似的代码，但这是我第一次遇到这种情况，所以我想知道为什么这段代码没有按预期工作

我希望程序会问我

“如果你同意，请键入“1”；如果你拒绝，请键入“0”。

21次，根据键入的

1或0，它会在变量“同意”或“拒绝”中添加一个，当然，在插入输入后会向我显示当前分数，如：

Who wants to play volleyball on sunday from 4pm to 6pm?
If more 12 or more people agree, I will book a field for us.
Type '1' if you agree - Type '0' if you decline.
1
1 YES vs. 0 NO
Type '1' if you agree - Type '0' if you decline.
0
1 YES vs. 1 NO
Type '1' if you agree - Type '0' if you decline.

# and so on...

但它却给了我：

Who wants to play volleyball on sunday from 4pm to 6pm?
If more 12 or more people agree, I will book a field for us.
Type '1' if you agree - Type '0' if you decline.
1
11
11
1 YES vs. 0 NO
Type '1' if you agree - Type '0' if you decline.
0
00
1 YES vs. 1 NO
Type '1' if you agree - Type '0' if you decline.
0
00
1 YES vs. 2 NO
Type '1' if you agree - Type '0' if you decline.
0
00
1 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
11
2 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
11
3 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
11
4 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
11
5 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
11
6 YES vs. 3 NO
Type '1' if you agree - Type '0' if you decline.
1
10
6 YES vs. 4 NO
Type '1' if you agree - Type '0' if you decline.
0
01
7 YES vs. 4 NO
Type '1' if you agree - Type '0' if you decline.
0
00
7 YES vs. 5 NO
Type '1' if you agree - Type '0' if you decline.
1
11
8 YES vs. 5 NO
Type '1' if you agree - Type '0' if you decline.
1
11
9 YES vs. 5 NO
Type '1' if you agree - Type '0' if you decline.
1
11
10 YES vs. 5 NO
Type '1' if you agree - Type '0' if you decline.
1
10
10 YES vs. 6 NO
Type '1' if you agree - Type '0' if you decline.
1
10
10 YES vs. 7 NO
Type '1' if you agree - Type '0' if you decline.
0
00
10 YES vs. 8 NO
Type '1' if you agree - Type '0' if you decline.
0
01
11 YES vs. 8 NO
Type '1' if you agree - Type '0' if you decline.
1
11
12 YES vs. 8 NO
Type '1' if you agree - Type '0' if you decline.
1
11
13 YES vs. 8 NO
Type '1' if you agree - Type '0' if you decline.
1
Cool,I am gonna book a field for us.SEE YOU ON SUNDAY at 4pm.LET'S GOOOOO!!!
Have a nice week until then !

如果您想知道为什么有时会出现
11、00、01或10
：
第二个数字是我必须再次插入作为输入的数字，第一个数字由程序打印出来

如果只有一个数字，
1或0
：
这是我的意见

正如你可能已经注意到的：
我先插入什么或第二个作为输入并不重要。因为某种原因，我不知道只有最后的输入才有意义

（我还没有弄清楚如何可视化（比如：斜体、粗体或颜色）您的代码，这就是为什么我必须在这里稍微解释一下代码的原因…）

我的代码：

print("Who wants to play volleyball on sunday from 4pm to 6pm?")
print("If more 12 or more people agree, I will book a field for us.")
print("Type '1' if you agree - Type '0' if you decline.")

answer = input()
z = int(input(answer))
agreements = 0
declines = 0

for i in range(0, 21):
    try:
        z = int(input(answer))
        if z == 1:
            agreements += 1
            print(agreements, "YES vs.", declines, "NO")
            print("Type '1' if you agree - Type '0' if you decline.")

            answer = input()
        elif z == 0:
            declines += 1
            print(agreements, "YES vs.", declines, "NO")
            print("Type '1' if you agree - Type '0' if you decline.")

            answer = input()
        else:
            raise
    except:
        print("Please type '1' or '0'.Try again.")

if agreements >= 12:
    print("Cool,I am gonna book a field for us."
          "SEE YOU ON SUNDAY at 4pm."
          "LET'S GOOOOO!!!")
else:
    print("Unfortunately we didn't manage to get enough people for booking a field."
          "Maybe next time! :)")

print("Have a nice week until then !")

不管我怎么看，我都找不到问题所在。
这将是一个愚蠢的错误

我提前表示歉意和感谢。
您要求的是更多的
输入（）。代码中的每个
input（）
都将打印括号中的内容（如果有），并等待用户输入：您

这就是为什么有比你想象的更多的投入——还有很多你认为不应该发生的产出。本质上，您是在提示自己输入一些文本，然后将这些文本用作消息，并打印为代码中下一个输入（）的提示

这：

下一个问题：即使给出了无效的输入，for循环也会增加，所以如果输入错误21次，也不会被询问21次“有效”

您可以通过不使用“数字”进行比较和使用函数询问（只要给出无效输入）来简化此操作：

def agreeToPlay():
    """Loop until either 0 or 1 is inputted - remove whitespsaces for convenience.
    Returns True on agreement, else False"""

    # loop with no end
    while True:
        choice = input("Type '1' if you agree - Type '0' if you decline:").strip()

        # only way out of the while is by inputing either 0 or 1:
        if choice in ["1","0"]:
            # only way to break out of the endless loop is valid input
            return choice == "1"  # return True if 1 else False

您可以在主程序中使用此功能：

print("Who wants to play volleyball on sunday from 4pm to 6pm?")
print("If more 12 or more people agree, I will book a field for us.")
print("Type '1' if you agree - Type '0' if you decline.")

agreements = 0
declines = 0
answers = []

# shortcut:
# while agreements + declines < 21 and agreements < 12  # ends as soon as 12 agreed

while agreements + declines < 21: 
    if agreeToPlay(): # ask until valid, then increment the correct one based
                      # on return of True / False
        agreements += 1
    else:
        declines +=1 

    print(agreements, "YES vs.", declines, "NO")

if agreements >= 12:
    print("Cool,I am gonna book a field for us."
          "SEE YOU ON SUNDAY at 4pm."
          "LET'S GOOOOO!!!")
else:
    print("Unfortunately we didn't manage to get enough people for booking a field."
          "Maybe next time! :)") 

print（“谁想在周日下午4点到6点打排球？”）
打印（“如果超过12人或更多人同意，我将为我们预订一个字段。”）
打印（“如果您同意，请键入“1”；如果您拒绝，请键入“0”。）
协议=0
下降=0
答案=[]
#快捷方式：
#协议+下降<21，协议<12#在12达成一致后立即结束
协议+下降<21:
如果agreeToPlay（）：#询问直到有效，然后基于
#返回真/假时
协议+=1
其他：
下降+=1
打印（协议“是”与“否”，拒绝“否”）
如果协议>=12：
打印（“酷，我要为我们预订一块地。”
“星期天下午4点见。”
“让我们加油！！！”）
其他：
print（“不幸的是，我们没有找到足够的人来预订场地。”
“也许下次吧！：）”）
你不应该做
z=int（输入（））
或
z=int（答案）
而不是
z=int（输入（答案））
？哦，是的，实际上使用
z=int（答案）
解决了我的问题。这是我的错，我有点加倍，使用输入和回答。非常感谢。但现在又出现了另一个问题。我的try and except现在似乎不起作用，因为每当我编写与0或1不同的内容时，程序都以
结束，请键入“1”或“0”。再试一次。
（循环范围的倍）
不幸的是，我们没有管理足够的人来预订字段。也许下次吧！：）在那之前祝你一周愉快不要求我再次插入通常应该插入的输入，对吗？哦，这很有意义。当我写这部分代码的时候我在想什么…嗯。哦，哇，这很有趣。它与我的atempt完全不同。谢谢你的见解和你的时间帮助我。我很感激。谢谢你，阿特纳先生！
print("Who wants to play volleyball on sunday from 4pm to 6pm?")
print("If more 12 or more people agree, I will book a field for us.")
print("Type '1' if you agree - Type '0' if you decline.")

agreements = 0
declines = 0
answers = []

# shortcut:
# while agreements + declines < 21 and agreements < 12  # ends as soon as 12 agreed

while agreements + declines < 21: 
    if agreeToPlay(): # ask until valid, then increment the correct one based
                      # on return of True / False
        agreements += 1
    else:
        declines +=1 

    print(agreements, "YES vs.", declines, "NO")

if agreements >= 12:
    print("Cool,I am gonna book a field for us."
          "SEE YOU ON SUNDAY at 4pm."
          "LET'S GOOOOO!!!")
else:
    print("Unfortunately we didn't manage to get enough people for booking a field."
          "Maybe next time! :)")