循环只运行一次答案

【问题标题】：the loop runs only once循环只运行一次
【发布时间】：2018-05-02 03:08:08
【问题描述】：

我的代码有问题。该循环仅运行一次，对应于用户输入的数字。感谢您的帮助。

#create an empty list and ask the user how many items they want to add  
# take the items the user entered and add it to the empty list.
print("enter the number of items")
x = int(input(">")) #take number of user input with the type int
print("thanks")
print("enter your food items")
fitems = [] # empty list
for i in range(x): #range to limit the number of iteration the user entered.
    items = input (">") #taking the items from the user 
    empty = ""
    if (items == empty):# checking if the user added nothing
        print("no items added")
    break
else:
    fitems.append(items) #add the list of items the user entered to the 
empty list
    print("hello here is your food items: \n",fitems) #output user items

【问题讨论】：

无论for 循环中发生什么，break 的缩进保证它在第一次迭代后中断。也许您打算将其缩进到 if 条件内。
您的break 命令位于for 循环内的主级别，因此它肯定会在第一次执行循环并结束时执行。尝试将break 语句再缩进一级，将其放在前面的if 语句中，看看是否能解决您的问题。

标签： python python-3.x

【解决方案1】：

这会起作用，因为你没有在 for 循环中缩进“break”

#create an empty list and ask the user how many items they want to add  
 # take the items the user entered and add it to the empty list.
 print("enter the number of items")
 x = int(input(">")) #take number of user input with the type int
 print("thanks")
 print("enter your food items")
 fitems = [] # empty list
 for i in range(x): #range to limit the number of iteration the user entered.
    items = input (">") #taking the items from the user 
    empty = ""
    if (items == empty):# checking if the user added nothing
       print("no items added")
       break #your error corrected
    else:
       fitems.append(items) #add the list of items the user entered to the 
  empty list
    print("hello here is your food items: \n",fitems) #output user items

【讨论】：

效果很好，但我希望打印语句在用户输入项目后打印所有内容，而不是打印一个并最终打印所有内容。

【解决方案2】：

我认为你的代码中唯一的问题是标识，所以你应该这样改变：

print("enter the number of items")
x = int(input(">")) #take number of user input with the type int
print("thanks")
print("enter your food items")
fitems = [] # empty list
for i in range(x): #range to limit the number of iteration the user ente red.
    items = input (">") #taking the items from the user 
    empty = ""
    if (items == empty):# checking if the user added nothing
        print("no items added")
        break
    else:
        fitems.append(items) #add the list of items the user entered to the 

print("hello here is your food items: \n",fitems)

【讨论】：

效果很好，但我希望在用户没有输入任何内容后不要打印最后一个打印语句

【解决方案3】：

您的代码中缺少一些缩进，并且还有一些其他错误。这是（我认为）正确的代码：

print("enter the number of items")
x = int(input(">"))
print("thanks")
print("enter your food items")
fitems = []
for i in range(x):
    items = input(">").strip() # there was a space before first bracket in your code, it's possible to have it there, but this is better
    # edit: added .strip(), it removes extra whitespace from beginning and end
    if items == "": # no need to create empty string variable, compare simply with empty string,
    # edit: you can also use 'if not items:', because empty string evaluates to false
        print("no items added")
        break # break only if nothing added, so indent break to be only executed if empty
    fitems.append(items) # you don't have to have else here, break ends the loop, so this is not executed then
print("hello here are your food items: \n",fitems) # remove 1 indent, print only after loop ended

【讨论】：

if (items == ""): 有不必要的括号，在大多数情况下，最好只检查一些虚假的东西。 if not items: 更 Pythonic。
效果很好，但我希望打印语句在用户输入项目后打印所有内容，而不是打印一个并最终打印所有内容。
@roganjosh 括号在那里是因为我最近在写一些 JS。至于not items，我爸爸（python 程序员）说我滥用了这种行为，我应该使用比较，因为它对任何阅读代码的人都是透明的。
@JonathanAkweteyOkine 我不确定我是否理解你...现在，代码在循环结束后打印所有项目。如果您希望它在每次用户添加项目时打印所有项目，请在最后一个 print 语句之前添加一个缩进级别（4 个空格），以便它进入循环并在每次迭代中执行。
@JonathanAkweteyOkine 我知道您是 Stack Overflow 的新手，所以这里有一个提示 - 如果这个（或任何其他，当然）答案解决了您的问题并且您认为它是最好的，请将其标记为已接受（打钩）。见What does it mean when an answer is "accepted"?

【解决方案4】：

首先，break 语句和 else 语句的意图不正确。此外，当您使用 .append() 时，您应该添加一个 i+=1 来循环 x 用户输入的内容。

这将起作用：

for i in range(x): #range to limit the number of iteration the user entered.
    items = input (">") #taking the items from the user
    empty = ""
    if (items == empty):# checking if the user added nothing
        print("no items added")
        break
    else:
        fitems.append(items) #add the list of items the user entered to the
        i += 1

【讨论】：

【解决方案5】：

如果没有输入，这将导致不打印：

print("enter the number of items")
x = int(input(">")) #take number of user input with the type int
print("thanks")
print("enter your food items")
fitems = [] # empty list
for i in range(x): #range to limit the number of iteration the user entered.
    items = input (">") #taking the items from the user 
    empty = ""
    if (items == empty):# checking if the user added nothing
        print("no items added")
        break
    else:
        fitems.append(items) #add the list of items the user entered to the\
                                                                empty list
if len(fitems)==0:
    pass
else:
    print("hello here is your food items: \n",fitems)

【讨论】：

【解决方案6】：

我冒昧地做了一些改变。试试这个：

# create an empty list and ask the user how many items they want to add  
# take the items the user entered and add it to the empty list.

fitems = []

# Define valid input for amount of items
valid = [str(i) for i in range(1,10)] # ["1","2"...."9"]

# If item is not in the list of valid, continue asking
while True:
    print("enter the number of items [1-9]")
    x = input(">")
    if x in valid:
        break

# Convert to int
itemamount = int(x)
print("thanks\nenter your food items")

# Now ask for inputs using while > 0 and remove itemamount in each loop
while itemamount > 0:
    while True:
        item = input (">")
        if item:
            break
        print("Blank input")
    fitems.append(item)
    itemamount-=1

print("hello here is your food items: ")
for ind,i in enumerate(fitems,1):
    print("{}. {}".format(ind,i))

【讨论】：