Python Elif语句在不应出现时出现'；t在列表中查找项目时_Python_Python 3.x

Python Elif语句在不应出现时出现'；t在列表中查找项目时

python python-3.x

Python Elif语句在不应出现时出现'；t在列表中查找项目时,python,python-3.x,Python,Python 3.x,因此，我不确定我在这里做错了什么，但我希望这样，如果他们键入“bloof”时出现拼写错误，它会说“未找到项”，但在这里使用elif语句，即使您键入“blood”，它仍然称为“未找到项” 如果您取出elif语句，键入“blood”将调用该项，但使用elif语句时，它总是显示“未找到项” 因为在输入if语句之前只检查第一项你应该这样做： res = "" for text2 in shopitemsF: if shopchoice in text2: res = text2

因此，我不确定我在这里做错了什么，但我希望这样，如果他们键入“bloof”时出现拼写错误，它会说“未找到项”，但在这里使用elif语句，即使您键入“blood”，它仍然称为“未找到项”

如果您取出elif语句，键入“blood”将调用该项，但使用elif语句时，它总是显示“未找到项”

因为在输入if语句之前只检查第一项

你应该这样做：

res = ""
for text2 in shopitemsF:
    if shopchoice in text2:
        res = text2
if res != "":
    print(res)
else:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")

# start by initializing the input:
shopchoice = input("Please choose an item from the shop by typing part of its name: ")
while True:  # main decision loop, repeat until we have a valid input
    shopchoice = shopchoice.title()  # capitalize the first letter
    found = None  # holds our found element, if any
    for text2 in shopitemsF:  # loop through all of the elements in the list
        if shopchoice in text2:  # check if the input is in the current element
            found = text2  # it is, store it in found for use bellow
            break  # item found, no need to search further, comment if you want the last item
    if found:  # we've found an element in the previous loop
        print(found)  # print it...
        break  # exit the main decision loop
    else:  # item was not found
        print("Item not found.")  
        shopchoice = input("Please pick another item? ")  # request for another input, repeat

就我个人而言，我会这样写：

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
"Bloodcursed Sword: 200 Damage, Cost: 950"]

item_not_found = True
while item_not_found:
    shopchoice = input("Please choose an item from the shop by typing part of its name: ")
    shopchoice = shopchoice.title()
    for text2 in shopitemsF:
        if shopchoice in text2:
            print(text2)
            item_not_found = False
            break
    if item_not_found:
        print("Item not found.")

for text2 in shopitemsF:
    if shopchoice in text2:
        print(text2)
        break

因此，每次找不到某个项时，它都会循环并重新提示用户。

因为在输入if语句之前，您只检查第一个项

你应该这样做：

res = ""
for text2 in shopitemsF:
    if shopchoice in text2:
        res = text2
if res != "":
    print(res)
else:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")

# start by initializing the input:
shopchoice = input("Please choose an item from the shop by typing part of its name: ")
while True:  # main decision loop, repeat until we have a valid input
    shopchoice = shopchoice.title()  # capitalize the first letter
    found = None  # holds our found element, if any
    for text2 in shopitemsF:  # loop through all of the elements in the list
        if shopchoice in text2:  # check if the input is in the current element
            found = text2  # it is, store it in found for use bellow
            break  # item found, no need to search further, comment if you want the last item
    if found:  # we've found an element in the previous loop
        print(found)  # print it...
        break  # exit the main decision loop
    else:  # item was not found
        print("Item not found.")  
        shopchoice = input("Please pick another item? ")  # request for another input, repeat

就我个人而言，我会这样写：

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
"Bloodcursed Sword: 200 Damage, Cost: 950"]

item_not_found = True
while item_not_found:
    shopchoice = input("Please choose an item from the shop by typing part of its name: ")
    shopchoice = shopchoice.title()
    for text2 in shopitemsF:
        if shopchoice in text2:
            print(text2)
            item_not_found = False
            break
    if item_not_found:
        print("Item not found.")

for text2 in shopitemsF:
    if shopchoice in text2:
        print(text2)
        break

因此，每次找不到某个项目时，它都会循环并重新向用户提示。

您的方法和下面的解决方案修复了您方法中的错误，但不会扩展

found = False
for text2 in shopitemsF:
    if shopchoice in text2:
        found = True
        break

if not found:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")
else:
    print("Item found")

我说的不是scalabe，因为如果您在

shopitemsF

中有N个项目，并且每个项目的平均长度为

，则此搜索将是O（NM）-可管理的小N和小M，但有数千条记录，它将非常缓慢。

您的方法和下面的解决方案，它修复了您方法中的错误，但不会扩展

found = False
for text2 in shopitemsF:
    if shopchoice in text2:
        found = True
        break

if not found:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")
else:
    print("Item found")

我说的不是scalabe，因为如果您在

shopitemsF

中有N个项目，并且每个项目的平均长度为

，则此搜索将是O（NM）-可管理的小N和小M，但有数千条记录，速度将非常慢。

您没有检查列表中所有可用项目的输入，就第一个。您需要遍历所有这些文件，以确保您的输入不在其中任何一个文件中。此外，如果输入无效，则需要返回到输入检查。比如：

res = ""
for text2 in shopitemsF:
    if shopchoice in text2:
        res = text2
if res != "":
    print(res)
else:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")

# start by initializing the input:
shopchoice = input("Please choose an item from the shop by typing part of its name: ")
while True:  # main decision loop, repeat until we have a valid input
    shopchoice = shopchoice.title()  # capitalize the first letter
    found = None  # holds our found element, if any
    for text2 in shopitemsF:  # loop through all of the elements in the list
        if shopchoice in text2:  # check if the input is in the current element
            found = text2  # it is, store it in found for use bellow
            break  # item found, no need to search further, comment if you want the last item
    if found:  # we've found an element in the previous loop
        print(found)  # print it...
        break  # exit the main decision loop
    else:  # item was not found
        print("Item not found.")  
        shopchoice = input("Please pick another item? ")  # request for another input, repeat

您没有检查列表中所有可用项的输入，只检查第一项。您需要循环遍历所有这些参数，以确保输入不在其中任何一个。此外，如果输入无效，则需要返回到输入检查。比如：

res = ""
for text2 in shopitemsF:
    if shopchoice in text2:
        res = text2
if res != "":
    print(res)
else:
    print("Item not found.")
    shopchoice = input("Please pick another item? ")

# start by initializing the input:
shopchoice = input("Please choose an item from the shop by typing part of its name: ")
while True:  # main decision loop, repeat until we have a valid input
    shopchoice = shopchoice.title()  # capitalize the first letter
    found = None  # holds our found element, if any
    for text2 in shopitemsF:  # loop through all of the elements in the list
        if shopchoice in text2:  # check if the input is in the current element
            found = text2  # it is, store it in found for use bellow
            break  # item found, no need to search further, comment if you want the last item
    if found:  # we've found an element in the previous loop
        print(found)  # print it...
        break  # exit the main decision loop
    else:  # item was not found
        print("Item not found.")  
        shopchoice = input("Please pick another item? ")  # request for another input, repeat

问题是您的for循环，您正在遍历整个列表并对每个列表应用检查。由于数组中有两个项目，并且您只测试一个项目，因此每次都会返回两个结果。要解决这个问题，只需在循环中添加一个

break

。看起来像这样：

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
"Bloodcursed Sword: 200 Damage, Cost: 950"]

item_not_found = True
while item_not_found:
    shopchoice = input("Please choose an item from the shop by typing part of its name: ")
    shopchoice = shopchoice.title()
    for text2 in shopitemsF:
        if shopchoice in text2:
            print(text2)
            item_not_found = False
            break
    if item_not_found:
        print("Item not found.")

for text2 in shopitemsF:
    if shopchoice in text2:
        print(text2)
        break

由于您不希望数组中不匹配的每个项目都出现“item not found”（项目未找到），因此我建议您创建一个变量来存储是否找到该项目，并结合while循环，在未找到该项目时再次触发商店，如果玩家不想购买，则创建一个结束循环的条目：

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
      "Bloodcursed Sword: 200 Damage, Cost: 950"]

shopchoice = input("Please choose an item from the shop by typing part of its name: ")

while True:
    shopchoice = shopchoice.title()

    if shopchoice == "End":
        break

    found = False
    for text2 in shopitemsF:
        if shopchoice in text2:
            found = text2
            break

    if found != False:
        print(found)
        break

    print("Item not found.")
    shopchoice = input("Please pick another item? ")

break

。看起来像这样：

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
"Bloodcursed Sword: 200 Damage, Cost: 950"]

item_not_found = True
while item_not_found:
    shopchoice = input("Please choose an item from the shop by typing part of its name: ")
    shopchoice = shopchoice.title()
    for text2 in shopitemsF:
        if shopchoice in text2:
            print(text2)
            item_not_found = False
            break
    if item_not_found:
        print("Item not found.")

for text2 in shopitemsF:
    if shopchoice in text2:
        print(text2)
        break

shopitemsF = ["Ghostblade: 150 Damage, Cost: 700", "Thunderblade: 120 Damage, Cost: 300",
      "Bloodcursed Sword: 200 Damage, Cost: 950"]

shopchoice = input("Please choose an item from the shop by typing part of its name: ")

while True:
    shopchoice = shopchoice.title()

    if shopchoice == "End":
        break

    found = False
    for text2 in shopitemsF:
        if shopchoice in text2:
            found = text2
            break

    if found != False:
        print(found)
        break

    print("Item not found.")
    shopchoice = input("Please pick another item? ")

我真的不知道你的意思。因此，在你的代码中输入for循环，检查第一个项目是否包含

血液

，然后告诉用户没有找到该项目。与通知用户之前检查所有项目不同。我真的不知道你的意思。因此，在代码中输入for循环，检查第一个项目是否包含

血液

，然后告诉用户未找到该项目。而不是在通知用户之前检查所有项目。您正在检查列表中的每个项目。如果第一个项目不包含血液，那么else语句将启动并要求您提供另一个输入。您要做的是在检查列表中的所有项目后检查是否找到匹配项。另外请注意，不需要elif。这只是一个else。哦，这是有道理的。我不认为它会这样做。你正在检查列表中的每一项。如果第一个项目不包含血液，那么else语句将启动并要求您提供另一个输入。您要做的是在检查列表中的所有项目后检查是否找到匹配项。另外请注意，不需要elif。这只是另一个例子。哦，这是有道理的，我不认为它是这样做的。是的，这起作用了，现在我知道为什么了，感谢@Thanassis。很高兴能提供帮助。是的，这起作用了，现在我知道为什么了，感谢@Thanassis。很高兴能提供帮助。