Python：从内部循环中断到外部while循环中的特定条件

Question

该程序运行良好，但如果我可以将内部循环中断到外部 while 循环中的特定条件会更好。 基本上，我不想再次运行整个循环，因为再次显示主菜单是多余的。 如果答案是“否”，则应该只是 go 到“查看并退出”选项。 如何从第 52 行（内循环）到第 84 行（外循环条件）中的 go？

def menu():   
    print('Main Menu')
    print('1) Chair')
    print('2) Table')
    print('3) Review and Exit')

def menu2():
    print('')
    print('1) Yes')
    print('2) No')

item1_count = 0
item2_count = 0
chair_count = 0
table_count = 0
loop = 1

while loop == 1:
    menu()
    while True:
        try:
            q = int(input('Choose an item: '))
        except ValueError:
            print('Choose 1-3.')
            continue
        else:
            break

    if q == 1:
        item1_count = item1_count + q
        while True:
            try:
                q2 = int(input('How many chairs? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        chair_count = chair_count + q2
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                # i want to go to line 83 here if i choose no.. how?
                # 'break' takes me back to the main loop (line 19)
                break

    elif q == 2:
        item2_count = item2_count + q
        while True:
            try:
                q4 = int(input('How many tables? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        table_count = table_count + q4
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                # i want to go to line 83 here if i choose no.. how?
                # 'break' takes me back to the main loop (line 19)
                break
    
    elif q == 3:
        print('You ordered', chair_count, 'chair(s) and', table_count, 'table(s).')
        break
    else:
        print('Choose 1-3.')
        continue

print(item1_count)
print(round(item2_count / 2))
print(chair_count)
print(table_count)

Answer 1

这是我的解决方案：

变化：

第 19 行：在开头初始化 q3
第 23 行：添加了 if 语句
第 32 行：添加了 elif 语句

def menu():
    print('Main Menu')
    print('1) Chair')
    print('2) Table')
    print('3) Review and Exit')


def menu2():
    print('')
    print('1) Yes')
    print('2) No')


item1_count = 0
item2_count = 0
chair_count = 0
table_count = 0
loop = 1
q3 = 0  # initialize q3 from the start because it can cause NameError on line 23 otherwise

while loop == 1:
    while True:
        if q3 != 2:  # create this condition, so if the user chose number 2 in the last question the code below will not execute
            menu()
            try:
                q = int(input('Choose an item: '))
            except ValueError:
                print('Choose 1-3.')
                continue
            else:
                break
        elif q3 == 2:  # and if the user chose 2 in the last question, we will set var q to 3 and break from the loop
            q = 3
            break

    if q == 1:
        item1_count = item1_count + q
        while True:
            try:
                q2 = int(input('How many chairs? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        chair_count = chair_count + q2
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                # i want to go to line 83 here if i choose no.. how?
                # 'break' takes me back to the main loop (line 19)
                break

    elif q == 2:
        item2_count = item2_count + q
        while True:
            try:
                q4 = int(input('How many tables? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        table_count = table_count + q4
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                # i want to go to line 83 here if i choose no.. how?
                # 'break' takes me back to the main loop (line 19)
                break

    elif q == 3:
        print('You ordered', chair_count, 'chair(s) and', table_count, 'table(s).')
        break
    else:
        print('Choose 1-3.')
        continue

print(item1_count)
print(round(item2_count / 2))
print(chair_count)
print(table_count)

推荐

老实说，这个程序很糟糕，它不可读并且很难修复错误。 我建议你尝试将这个程序拆分成函数，这样代码会更易读，更专业。

你可以阅读这篇文章有一个更好的理解。

Answer 2

我不确定是否要根据外循环条件打破内循环，除非我们在内循环的每次迭代中检查条件。

但我希望下面的修改会产生你想要的结果：

def menu():   
    print('Main Menu')
    print('1) Chair')
    print('2) Table')
    print('3) Review and Exit')

def menu2():
    print('')
    print('1) Yes')
    print('2) No')

item1_count = 0
item2_count = 0
chair_count = 0
table_count = 0
loop = 1
stopped=False

while loop == 1:
   
    while stopped==False:
        menu()
        try:
            q = int(input('Choose an item: '))
        except ValueError:
            print('Choose 1-3.')
            continue
        else:
            break

    if q == 1:
        item1_count = item1_count + q
        while True:
            try:
                q2 = int(input('How many chairs? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        chair_count = chair_count + q2
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                stopped=True
                q=3
                break
                

    elif q == 2:
        item2_count = item2_count + q
        while True:
            try:
                q4 = int(input('How many tables? '))
            except ValueError:
                print('Type a number.')
                continue
            else:
                break
        table_count = table_count + q4
        menu2()
        while True:
            try:
                q3 = int(input('Would you like anything else? '))
            except ValueError:
                print('Choose 1 or 2.')
                continue
            if q3 == 1:
                print('yes')
                break
            elif q3 == 2:
                print('no.')
                q=3
                stopped=True
                break
               
    
    elif q == 3:
        print('You ordered', chair_count, 'chair(s) and', table_count, 'table(s).')
        break
    else:
        print('Choose 1-3.')
        continue

print(item1_count)
print(round(item2_count / 2))
print(chair_count)
print(table_count)

在这里，我们在全局级别定义了一个名为stopped的变量，它告诉用户想要停止运行循环，并且当用户想要退出时我们将其设置为True 。 如果是真的，我们将停止征求意见。 此外，我们将值 3 分配给变量q以将 if 语句重定向到打印评论。 请注意，我们将调用menu() function 转移到了 while 循环内部。

and statements.但是尝试用更少的和语句来设计流程。 可以针对上述情况进行。

我希望这有帮助。