对于一个向前的后藤，你可以添加:

while True:
  if some condition:
    break
  #... extra code
  break # force code to exit. Needed at end of while loop
#... continues here

但这只适用于简单的场景(即嵌套这些会让你陷入混乱)

我认为while循环是“goto_Statement”的替代。因为3.6之后，goto循环不再工作了。我还写了一个while循环的例子。

str1 = "stop"
while str1 == "back":
    var1 = int(input(" Enter Ist Number: "))
    var2 = int(input(" Enter 2nd Number: "))
    var3 = print("""  What is your next operation
                      For Addition   Press And Enter : 'A'
                      For Muliplt    Press And Enter : 'M'
                      For Division   Press And Enter : 'D'
                      For Subtaction Press And Enter : 'S' """)

    var4 = str(input("For operation press any number : "))
    if(var1 == 45) and (var2 == 3):
        print("555")
    elif(var1 == 56) and (var2 == 9):
        print("77")
    elif(var1 == 56) and (var2 == 6):
        print("4")
    else:
        if(var4 == "A" or "a"):
            print(var1 + var2)
        if(var4 == "M" or "m"):
            print(var1 * var2)
        if(var4 == "D" or "d"):
            print(var1 / var2)
        if(var4 == "S" or "s"):
            print(var1 - var2)

    print("if you want to continue then type  'stop'")

    str1 = input()
print("Strt again")    

早在2007年，PEP 3136就提出了中断和继续的标签，但被拒绝了。提案的动机部分说明了在Python中模仿带标签的break的几种常见方法(虽然不太优雅)。

我想要同样的答案，我不想使用goto。所以我使用了下面的例子(来自learnpythonthehardway)

def sample():
    print "This room is full of gold how much do you want?"
    choice = raw_input("> ")
    how_much = int(choice)
    if "0" in choice or "1" in choice:
        check(how_much)
    else:
        print "Enter a number with 0 or 1"
        sample()

def check(n):
    if n < 150:
        print "You are not greedy, you win"
        exit(0)
    else:
        print "You are nuts!"
        exit(0)

不，Python不支持标签和goto。它是一种高度结构化的编程语言。

我有自己的做法。 我使用单独的python脚本。

如果我想要循环:

file1.py

print("test test")
execfile("file2.py")
a = a + 1

file2.py

print(a)
if a == 10:
   execfile("file3.py")
else:
   execfile("file1.py")

file3.py

print(a + " equals 10")

(注意:此技术仅适用于Python 2。x版本)