Question

我的代码获得了一些输入，并最终输出了一些结果。我已经将a或ice cream函数的含义与shortcutA()关联在一起，将/的含义与产品分离，将everything else的含义称为unrecognized。此后，将要求用户输入每种产品的数量，并打印出他所输入内容的总和，并对打印输出进行一些修改。

sentences = str(input("please enter the password :"))

for i in sentences:
  if i == "/":
    howMuchOrder = input("please enter the order :")
  else:
    continue

class meaning():
  def shortcutA (self):
    global sentences
    print ("ice cream")
    for i in sentences:
        if i != "/":
          print ("sweet ice") 
        elif i =="/":
          print ('order is ' + str(int(howMuchOrder)))

def main():
    m = meaning()
    print_flag = False
    for i in sentences :
        if i in ['a', 'b', '/'] and not print_flag:
          print("your code is: ")
          print_flag = True
        if i == "a" :
          m.shortcutA()
        elif i == "/":
          break
        else :
             print ("unrecognized")


if __name__ == "__main__":
  main()

说输入只是单词 a ，那么结果将是：       您的代码是：       冰淇淋甜冰

假设输入只是单词 a / ，那么结果将是：       您的代码是：       冰淇淋订单是20块甜冰

如果输入为 a / a / ，并且订单均为20（两者），则预期结果应为：

your code is: ice cream order is 20 sweet ice ice cream order is 20 sweet ice

Answer 1

据我了解，您对待

a快捷方式为ice cream
whatever else作为unrecognized

然后为您要打印的shortcutA ice cream order is <order_amount> sweet ice

基于这些理解，鉴于给定a/a/，您期望ice cream order is 20 sweet ice ice cream order is 20 sweet ice我已经修改了部分逻辑和部分内容，但并没有全部内容向您展示如何实现所需的输出（可以在python 2.x中工作，最后看看如何使其在python 3.x中工作）：

# now input() works in same way on either python 2.x or 3.x
try:
    input = raw_input
except NameError:
    pass

sentences = str(input("please enter the password :"))

# split sentences by '/', remove empty values, put result in a list
listOrderName = list(filter(None, sentences.split('/')))
totalOrder = sentences.count('/')
listOrderAmount = [0]*totalOrder

if totalOrder > 0:
    for index, i in enumerate(listOrderAmount):
        howMuchOrder = input("please enter the order :")
        listOrderAmount[index] = howMuchOrder

#Exmple of what are your two input
print(listOrderName) #can be deleted
print(listOrderAmount) #can be deleted

class meaning():
  def shortcutA (self, position):
    global listOrderAmount, totalOrder

    print("ice cream"),

    if totalOrder > 0:
        totalOrder = totalOrder - 1
        amount = listOrderAmount[position]
        print('order is ' + str(int(amount))),

    print("sweet ice")

  def shortcutUnrecognized(self):
    print("unrecognized")

def main():
    m = meaning()
    print_flag = False
    position = -1

    global listOrderName
    for order in listOrderName :
        position += 1
        if order in ['a', 'b'] and not print_flag:
          print("your code is: ")
          print_flag = True
        if order == "a" :
          m.shortcutA(position)
        else :
          m.shortcutUnrecognized()  


if __name__ == "__main__":
  main()

用法示例：

#please enter the password :a/a/
#please enter the order :20
#please enter the order :20
#['a', 'a']
#['20', '20']
#your code is: 
#ice cream order is 20 sweet ice
#ice cream order is 20 sweet ice

另一个用法示例：

#please enter the password :a/b
#please enter the order :20
#please enter the order :20
#['a', 'b']
#['20', '20']
#your code is: 
#ice cream order is 20 sweet ice
#unrecognized

另一个用法示例：

#please enter the password :a/a
#please enter the order :20
#['a', 'a']
#['20']
#your code is: 
#ice cream order is 20 sweet ice
#ice cream sweet ice

您可以删除对您无用的照片（例如print(listOrderName)和print(listOrderAmount)。

python 2.x 例如，在print("ice cream"),中，,的作用是在打印后不放置EOL（行尾或换行），因此我们在同一行中进行了一些打印，而在其他行中进行了其他打印。这适用于python 2.x

python 3.x 要实现与print(..),中的python 3.x相同的行为，您可以将print("ice cream"),替换为print("ice cream", end =" ")。因此，实质上，要使脚本以与python 3.x的打印部分相同的方式工作，请更改以下内容：

print("ice cream"),
print('order is ' + str(int(amount))),

对此：

print("ice cream", end =" ")
print('order is ' + str(int(amount)), end =" ")

无法达到我的for循环中预期的打印结果

1 个答案: