你如何在python 3.x中使用或运算？

Posted 2023-02-27

def members_ID():
meme = False
membersID = str(input("enter members ID"))
while (meme) == False:
    meme = True
    if len(membersID) == 4:
        if not (membersID)[0] == "0" or "1" or "2" or "3" or "4" or "5" or "6" or "7" or "8" or "9":
            meme = False
        if not (membersID)[1] == "0" or "1" or "2" or "3" or "4" or "5" or "6" or "7" or "8" or "9":
            meme = False
        if not (membersID)[2] == "0" or "1" or "2" or "3" or "4" or "5" or "6" or "7" or "8" or "9":
            meme = False
        if not (membersID)[3] == "0" or "1" or "2" or "3" or "4" or "5" or "6" or "7" or "8" or "9":
            meme = False

    else:
        meme = False
        print("members ID must be in the format 1111")
        membersID = str(input("enter members ID"))
if meme == True:
    print(membersID)
elif (meme) == False:
    print("members ID must be in the format 1111")

如果它不是四个它可以工作，但如果是，那么程序将停止而不工作？

它可能与or运营商有关，因为当我拿出它们并输入4位数字所需的信息然后它可以工作。

def members_ID():
        meme = False
        membersID = str(input("enter members ID"))
        while (meme) == False:
            meme = True
            if len(membersID) == 4:
                if not (membersID)[0] == "0":
                    meme = False
                if not (membersID)[1] == "0":
                    meme = False
                if not (membersID)[2] == "0":
                    meme = False
                if not (membersID)[3] == "0":
                     meme = False

            else:
            meme = False
            print("members ID must be in the format 1111")
            membersID = str(input("enter members ID"))
    if meme == True:
        print(membersID)
    elif (meme) == False:
        print("members ID must be in the format 1111")

但是如果输入的信息是错误的。例如，ssss将使程序停止工作，但0000将（如果0000是所需的信息）！

答案

你不能以这种方式使用or。下列：

not (membersID)[0] == "0" or "1"

是true，无论是not membersID[0] == "0"还是字符串"1"都是非空的，它始终是。因此，if声明始终是true。

你的意图应该写成：

not (membersID[0] == "0" or membersID[0] == "1" or membersID[0] == "2") # etc. for 3..9

为了避免重复这么多，你可以这样做：

membersID[0] not in ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]

另一答案

or-operator位于两个语句或子句之间，并返回非None，False或等效的第一部分（如空字符串，列表等）

看这个例子：

a = 'a'
b = 'b'

print(a == 'b' or 'a')
# 'a'
print(a == 'a' or 'b')
# True

首先，它打印出值'a'，因为a == 'b'为False，'a'不为空，False或None。

其次，它打印出值True，因为a == 'a'为True。

如果您希望值为None，则这很方便：

x = some_variable or 5

这里，当x为None，False（或空）时，some_variable为5。

所以在你的链中：

如果不是（membersID）[0] ==“0”或“1”或“2”或“3”或“4”或“5”或“6”或“7”或“8”或“9”： meme =假

我希望meme = False可以在任何情况下运行，因为如果你将它重新标记为：

a = not (membersID)[0] == "0"
b = "1"
c = "2"
#...

然后你的if语句就是

if a or b or c  # or d ...
    meme = False

只有第一个条款，a将有机会不成真。

换句话说，如果if something具有价值，那么做something将是真的：

if "2":
    print('This is a tautology')

此外，如果ors链中的任何子句都没有被解析为True，它将返回最后一个子句：

x = None or False or None
print(x is None)
# True
print(type(x))
# NoneType
x = None or False
print(x)
# False

另一答案

这是我根据Thijs van dien的答案完成的答案。 （谢谢）

def members_ID():
    meme = False
    membersID = input("enter members ID")
    while (meme) == False:
        meme = True
        if len(membersID) == 4:
            if membersID[0] not in ["0","1","2","3","4","5","6","7","8","9"]:
                meme = False
            if membersID[1] not in ["0","1","2","3","4","5","6","7","8","9"]:
                meme = False
            if membersID[2] not in ["0","1","2","3","4","5","6","7","8","9"]:
                meme = False
            if membersID[3] not in ["0","1","2","3","4","5","6","7","8","9"]:
                meme = False
            if meme == True:
                print(membersID)
            elif meme == False:
                print("enter in the format 1111")
                membersID = input("enter members ID")

        else:
            meme = False
            print("members ID must be in the format 1111")
            membersID = input("enter members ID")

另一答案

membersID = str(input("enter members ID"))

input（）已经返回一个字符串。在此行中，您尝试在输入字符串时将字符串转换为字符串。因此，它会引发错误。纠正它：

membersID = input("enter members ID")