如何通过在python中创建条件来打破循环

Question

lst = []
while True:
    try:
        arr = int(input("Enter number of elements: "))
        if arr == "Quit":
            break
    except ValueError:
        print("Invalid Input")
        continue
    else:
        break
while True:
    try:
        for i in range(0, arr):
            ele = int(input("Enter the elements:"))
            lst.append(ele)
            print(lst)
    except ValueError:
        print("Invalid Input")
        continue
    else:
        break

如何通过专门输入条件项来创建条件以在循环中的任何点退出程序？ 就像当它要求我输入一个元素时一样，但我想在那个时候通过输入来破坏程序 “放弃”。 如何在这段代码中做到这一点？ （学习者）

Answer 1

使用sys.exit() 退出程序完全。使用break 退出您所在的特定循环（您已经使用过）。

其他说明：

1. 您正在执行arr = int(input(...))，但您想接受"Quit" 的输入，这是一个字符串。因此，如果用户输入“退出”，它会引发ValueError 并且循环继续。所以首先检查“退出”，如果它没有退出，然后使用 try-block 转换为int

   • 同样适用于向用户询问列表中元素的第二个循环

2. 顺便说一句，您的第二个循环的 while True 循环应该在获取每个元素的 for 循环内。

lst = []
while True:
    arr = input("Enter number of elements: ")
    if arr == "Quit":
        sys.exit()  # will exit completely
    try:
        arr = int(arr)  # check for int in the try-block, error raised here, if not int
        break           # can put break here instead of in else, any is okay
    except ValueError:
        print("Invalid Input")
        # continue not needed here, since it loops infinitely by default


for i in range(0, arr):
    while True:
        ele = input("Enter the elements:")
        if ele == "Quit":
            sys.exit()  # will exit completely
        try:
            ele = int(ele)  # check for int in the try-block, error raised here, if not int
            lst.append(ele)
            break           # breaks out of `while` loop, not `for` loop; good
        except ValueError:
            print("Invalid Input")
            # continue not needed here, since it loops infinitely by default

print(lst)  # print the full list after all the inputs

重复输入的模式：

• 要求用户输入
• 如果输入是“退出”，则完全退出
• 否则，将输入转换为 int

所以这可以放入一个函数中，你可以在两个地方调用它：

def int_or_quit(msg):
    """`msg` is the message you want to show at input"""
    while True:
        item = input(msg)
        if item == "Quit":
            sys.quit()  # will exit completely
        try:
            return int(item)  # try converting to int and return
        except ValueError:
            print("Invalid Input")
            # repeats by default

# use the function above in your two code blocks, which are now simplified:
lst = []
arr = int_or_quit("Enter number of elements: ")
# program exits before if user "Quit", next part won't execute

for i in range(0, arr):
    ele = int_or_quit("Enter an element:")
    lst.append(ele)

print(lst)

Answer 2

我可以看到您的代码 sn-p 中有 2 个块。我建议将它们组织成不同的函数，其中函数名称推断函数目标（而不是 cmets）：

def get_list_size():...
def fill_list():...

接下来要做的是决定函数的返回值：

from typing import List, Optional

def get_list_size() -> Optional[int]:
"""
:returns: number of desired elements of list or None, if "Quit" was pressed
:raises: ValueError exception, if non integer value was pressed
"""

def fill_list(lst_size: int) -> List[int]:
"""
returns: filled list
throws: ValueError exception, if non integer value was pressed
"""

在这里你可以看到，如何使用 Optional 类型提示将条件逻辑添加到程序中。
Lats 要做的就是用轻量级代码填充你的函数：

def get_list_size() -> Optional[int]:
    """
    :returns: number of desired elements of list or None, if "Quit" was pressed
    :raises: ValueError exception, if non integer value was pressed
    """
    while True:
        try:
            input_value = input("Enter number of elements: ")

            # Pay attention: you have to check string input_value, not    int(input_value).
            # Checking int(input_value) would raise exception before 'Quit' validation.
            if input_value == "Quit":
                return None

            return int(input_value)

            # Could be rewritten as trinary if
            # return None if input_value == "Quit" else int(input_value)
        except ValueError:
            print("Invalid Input")
            # continue <- Not needed here


def fill_list(lst_size: int) -> List[int]:
    """
    returns: filled list
    throws: ValueError exception, if non integer value was pressed
    """
    lst = []
    while len(lst) < lst_size:
        try:
            ele = int(input("Enter the elements:"))
            lst.append(ele)
            print(lst)
        except ValueError:
            print("Invalid Input")
        
    return lst

注意：在 fill_list 函数中，你必须尝试/排除每个元素，所以 while 循环被移到了外面。
最后要做的是运行这些函数：

desired_list = []
desired_list_size = get_list_size()
if desired_list_size is not None:
    desired_list = fill_list(desired_list_size)

Answer 3

def one():
    message = ""
    while message != "Quit":
        message = input("Type in Order > ")
        if message == "Run":
            print("Run some Code")
            message = input("What do u want to run ? > ")
            if message == "Exel":
                print("Runing Exel")
            else:
                pass
         else:
             pass

while True:
    one()

在这里，您可以传入在您通过名称调用它时将运行的函数。如果你写退出它将重新开始。如果你写错了，它会重新开始。如果你愿意，你可以为此做一个 FailExeption。