将字典项作为字符串返回时出错？(python)
将字典项作为字符串返回时出错？(python)
提问于 2021-12-09 18:49:56
我有一个类，它有一个空的字典属性：
class Library:
def __init__(self,library):
self.library={}
def addSong(self,title,artist,genre,playCount):
for i in self.library.keys():
if i == title:
x=self.library.get(title)
x[2]= x[2]+1
else:
self.library[title]=[artist,genre,playCount]
def song2String(self,title):
x=self.library.get(title)
return f"{x[0]} {title} ({x[1]}), {x[2]}" 当我这么做的时候：
m1= Library({})
m1.addSong("Saturday Night's Alright for Fighting","Elton John","Rock",22)代码正常运行，项被添加到m1字典中。但当我打字时：
print(m1.song2String("Saturday Night's Alright for Fighting"))我收到以下错误消息：
return f"str{x[0]} str{title} (str{x[1]}), str{x[2]}"
TypeError: 'NoneType' object is not subscriptable我的语法错误是什么？
回答 3
Stack Overflow用户
回答已采纳
发布于 2021-12-09 18:54:44
您在library键上迭代：for i in self.library.keys()，但是由于没有，什么也不做，使其工作的最小更改是
def addSong(self, title, artist, genre, playCount):
for i in self.library.keys():
if i == title:
x = self.library.get(title)
x[2] = x[2] + 1
break
else:
self.library[title] = [artist, genre, playCount]但是最好的方法就是测试标题是否存在，然后做正确的事情。
def addSong(self, title, artist, genre, playCount):
if title in self.library:
self.library[title][2] += playCount
else:
self.library[title] = [artist, genre, playCount]Stack Overflow用户
发布于 2021-12-09 18:55:48
当前代码不返回任何内容，这是导致错误的原因。我不知道你为什么要用钥匙插入你的歌，直接用字典就行了：
class Library:
def __init__(self,library):
self.library={}
def addSong(self,title,artist,genre,playCount):
if title in self.library: ## changed code here and below
x=self.library.get(title)
x[2]= x[2]+1
else:
self.library[title]=[artist,genre,playCount]
def song2String(self,title):
x=self.library.get(title)
return f"{x[0]} {title} ({x[1]}), {x[2]}" 产出：
m1= Library({})
m1.addSong("Saturday Night's Alright for Fighting","Elton John","Rock",22)
print(m1.library)
# {"Saturday Night's Alright for Fighting": ['Elton John', 'Rock', 22]}
print(m1.song2String("Saturday Night's Alright for Fighting"))
# Elton John Saturday Night's Alright for Fighting (Rock), 22Stack Overflow用户
发布于 2021-12-09 19:28:27
，我想这是你所期望的。在这里的加载项函数中，我首先检查的是字典中可用的任何键。如果是真的，那就迭代，做你想做的事。否则，将第一个字典项添加到字典中。在songs2string方法中，首先要识别字典是如何工作的。在这种情况下，键是字符串，值是字典中的列表。所以你可以用字典返回你想要的东西。
class Library:
def __init__(self,library=dict()):
self.library=library
def addSong(self,title,artist,genre,playCount):
if len(self.library.keys())>0:
for i in self.library.keys():
if i == title:
x=self.library.get(title)
x[2]= x[2]+1
else:
self.library[title]=[artist,genre,playCount]
# print(self.library)
def song2String(self,title):
x=self.library
print(x)
return x[title][0], x[title][1], x[title][2]
m1= Library()
m1.addSong("Saturday Night's Alright for Fighting","Elton John","Rock",22)
print(m1.song2String("Saturday Night's Alright for Fighting"))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/70295196
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