你可以使用一个标志来打破循环:

if found:
  break

这里，'found'是标志，你最初将它设置为False，然后在循环中使用这段代码。

found = False
for table_height in range(500):
  if found:
    break

下面是三个for循环的完整代码:

found = False
for table_height in range(500):
  if found:
    break
  for cat_height in range(500):
    if found:
      break

    for tort_height in range(500):
      equation1 = table_height + cat_height == tort_height + 170
      equation2 = table_height + tort_height == cat_height + 130
      if equation1 and equation2:
        print('table', table_height, '  cat', cat_height, '  tortoise', tort_height)
        found = True
        break

在这段代码中，如果等式1和等式2为True，它将设置'found'标志为True，并跳出最里面的for循环，它也将跳出其他两个for循环，因为'found'为True。

首先，普通的逻辑是有用的。

如果由于某种原因，终止条件无法确定，例外是一个后备计划。

class GetOutOfLoop( Exception ):
    pass

try:
    done= False
    while not done:
        isok= False
        while not (done or isok):
            ok = get_input("Is this ok? (y/n)")
            if ok in ("y", "Y") or ok in ("n", "N") : 
                done= True # probably better
                raise GetOutOfLoop
        # other stuff
except GetOutOfLoop:
    pass

对于这个特定的示例，可能不需要异常。

另一方面，在字符模式应用程序中，我们经常有“Y”、“N”和“Q”选项。对于"Q"选项，我们要立即退出。这是比较特殊的。

首先，您还可以考虑将获取和验证输入的过程作为一个函数;在该函数中，如果值正确，则返回值，如果不正确，则继续在while循环中旋转。这本质上避免了您解决的问题，并且通常可以应用于更一般的情况(打破多个循环)。如果你一定要在代码中保留这个结构，并且真的不想处理记帐布尔值……

你也可以用下面的方式使用goto(在这里使用一个愚人节模块):

#import the stuff
from goto import goto, label

while True:
    #snip: print out current state
    while True:
        ok = get_input("Is this ok? (y/n)")
        if ok == "y" or ok == "Y": goto .breakall
        if ok == "n" or ok == "N": break
    #do more processing with menus and stuff
label .breakall

我知道，我知道，“你不应该使用goto”之类的，但它在这种奇怪的情况下很管用。

以下是一个非常简短的版本: 创建名为break_out_nested.py的文件

import itertools
import sys

it = sys.modules[__name__] # this allows us to share variables with break_out_nested.py when we import it 


def bol(*args):
    condi = args[-1] # the condition function
    i = args[:-1] # all iterables 
    for p in itertools.product(*i): # itertools.product creates the nested loop
        if condi(): # if the condition is True, we return 
            return
        yield p # if not, we yield the result 

现在你只需要几行就可以打破嵌套的循环(数据来自Rafiq的例子)

from break_out_nested import it, bol # import what we have just created

# you need to create new variables as attributes of it,
# because break_out_nested has only access to these variables
it.i, it.j, it.k = 1, 1, 1
# the break condition
def cond(): return it.i % 3 == 0 and it.j % 3 == 0 and it.k % 3 == 0

# The condition will be checked in each loop 
for it.i, it.j, it.k in bol(range(1, 6, 1), range(1, 11, 2, ), range(1, 21, 4), cond):
    print(it.i, it.j, it.k)

更多的例子:

def cond(): return it.i + it.j + it.k == 777

it.i, it.j, it.k = 0, 0, 0
for it.i, it.j, it.k in bol(range(100), range(1000), range(10000), cond):
    print(it.i, it.j, it.k)




def cond(): return it.i + it.j + it.k >= 100000

it.i, it.j, it.k = 0, 0, 0
# you dont have to use it.i, it.j, it.k as the loop variables, you can
# use anything you want, but you have to update the variables somewhere
for i, j, k in bol(range(100), range(1000), range(10000), cond):
    it.i, it.j, it.k = i * 10, j * 100, k * 100
    print(it.i, it.j, it.k)

我解决这个问题的方法是定义一个变量，该变量被引用来确定你是否进入下一关。在这个例子中，这个变量叫做shouldbreak。

Variable_That_Counts_To_Three=1
while 1==1:
    shouldbreak='no'
    Variable_That_Counts_To_Five=0
    while 2==2:
        Variable_That_Counts_To_Five+=1
        print(Variable_That_Counts_To_Five)
        if Variable_That_Counts_To_Five == 5:
            if Variable_That_Counts_To_Three == 3:
                shouldbreak='yes'
            break
    print('Three Counter = ' + str(Variable_That_Counts_To_Three))
    Variable_That_Counts_To_Three+=1
    if shouldbreak == 'yes':
        break

print('''
This breaks out of two loops!''')

这为您希望程序中断的确切方式提供了大量控制，允许您选择何时中断以及下降多少层。

# this version uses a level counter to choose how far to break out

break_levels = 0
while True:
    # snip: print out current state
    while True:
        ok = get_input("Is this ok? (y/n)")
        if ok == "y" or ok == "Y":
            break_levels = 1        # how far nested, excluding this break
            break
        if ok == "n" or ok == "N":
            break                   # normal break
    if break_levels:
        break_levels -= 1
        break                       # pop another level
if break_levels:
    break_levels -= 1
    break

# ...and so on