不，有另一种方法来实现goto语句

class id:
     def data1(self):
        name=[]
        age=[]   
        n=1
        while n>0:
            print("1. for enter data")
            print("2. update list")
            print("3. show data")
            print("choose what you want to do ?")
            ch=int(input("enter your choice"))
            if ch==1:    
                n=int(input("how many elemet you want to enter="))
                for i in range(n):
                    name.append(input("NAME "))
                    age.append(int(input("age "))) 
            elif ch==2:
                name.append(input("NAME "))
                age.append(int(input("age ")))
            elif ch==3:
                try:
                    if name==None:
                        print("empty list")
                    else:
                        print("name \t age")
                        for i in range(n):
                            print(name[i]," \t ",age[i])
                        break
                except:
                    print("list is empty")
            print("do want to continue y or n")
            ch1=input()
            if ch1=="y":
                n=n+1
            else:
                print("name \t age")
                for i in range(n):
                    print(name[i]," \t ",age[i])
                n=-1
p1=id()
p1.data1()  

使用评论中@bobince的建议来回答@ascobol的问题:

for i in range(5000):
    for j in range(3000):
        if should_terminate_the_loop:
           break
    else: 
        continue # no break encountered
    break

else块的缩进是正确的。代码在循环Python语法后使用模糊的else。参见为什么python在for和while循环之后使用'else' ?

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")    

为了代替python的goto等效，我以以下方式使用break语句快速测试我的代码。这里假设您有结构化的代码库。测试变量在函数开始时初始化，我只是将“If test: break”块移动到我想测试的嵌套If -then块或循环的末尾，在代码末尾修改返回变量以反映我正在测试的块或循环变量。

def x:
  test = True
  If y:
     # some code
     If test:
            break
  return something