我对编程比较陌生,我有一个小问题,为ubuntu(linux)写一个python等效的snip for spotify 不知何故,我可以正确编码标题,但无法以相同的方式编码艺术家
当我尝试以相同的方式对艺术家进行编码时,我得到了这个:
File "./songfinder.py", line 11, in currentplaying
artiststr = str((metadata['xesam:artist']).encode('utf-8'))
AttributeError: 'dbus.Array' object has no attribute 'encode'
然而标题完全相同,而且工作正常。
到目前为止,代码正在运行,但例如\ xd8而不是Ø,类似:
import dbus
session_bus = dbus.SessionBus()
spotify_bus = session_bus.get_object("org.mpris.MediaPlayer2.spotify", "/org/mpris/MediaPlayer2")
spotify_properties = dbus.Interface(spotify_bus, "org.freedesktop.DBus.Properties")
def currentplaying():
metadata = spotify_properties.Get("org.mpris.MediaPlayer2.Player", "Metadata")
title = str((metadata['xesam:title']).encode('utf-8'))
artiststr = str((metadata['xesam:artist']))
if ("dbus.string" in artiststr.lower()):
artists = artiststr.split("(u")
artist = artists[1]
artists = artist.split(")],")
artist = artists[0]
artist = artist.replace("(u", "")
else:
artist = "'unknown'"
artist = (artist.replace("'",""))
playing = (artist + " - " + title + " ")
return playing
#save playing to file.txt
相关的qna: Replace non-ascii chars from a unicode string in Python
为什么它不能解决我的问题:我想打印/保存实际字符,而不是用类似字符替换它
答案 0 :(得分:0)
你得到的错误不是关于unicode,而是关于错误的类型。 Python抱怨你试图从数组对象调用字符串方法encode
。哪个没有这个方法。
我要尝试的第一个是删除多余的括号,这样就得到像这样的艺术家:artiststr = str(metadata['xesam:artist'])
。
但我不确定这会奏效。如果它不起作用,您需要找出元数据的类型[' xesam:artist']。看起来它不是字符串,而是数组。因此,您需要修复用metadata['xesam:artist']
填充数据的代码。您可以尝试使用调试器或print()
函数来查找metadata['xesam:artist']
的内容。或者也提供相关代码。
答案 1 :(得分:0)
最终节目,如果您愿意,请随时使用:
import time
import dbus
session_bus = dbus.SessionBus()
spotify_bus = session_bus.get_object("org.mpris.MediaPlayer2.spotify", "/org/mpris/MediaPlayer2")
spotify_properties = dbus.Interface(spotify_bus, "org.freedesktop.DBus.Properties")
def currentplaying():
metadata = spotify_properties.Get("org.mpris.MediaPlayer2.Player", "Metadata")
title = str((metadata['xesam:title']).encode('utf-8'))
artiststr = str((metadata['xesam:artist'])[0].encode('utf-8'))
artist = artiststr
playing = (artist + " - " + title + " ")
return playing
while True:
filetxt = open("/home/USER/Desktop/currentsongspotify.txt", "r")
oldtitle = filetxt.read()
filetxt.close()
newtitle = str(currentplaying())
if(newtitle == oldtitle):
time.sleep(1)
else:
filetxt = open("/home/USER/Desktop/currentsongspotify.txt", "w") #save newtitle to file, overwriting existing data
filetxt.write(str(newtitle))
print("new file saved: " + newtitle)
答案 2 :(得分:0)
查看您的问题metadata
至少包含类似于Unicode字符串的内容。艺术家领域似乎是与艺术家一起开始的某种可迭代的。这样的事情(随意发布实际的元数据内容):
metadata = {'xesam:title':u'title','xesam:artist':[u'artist']}
在title
分配行中,str
是不必要的,因为编码Unicode字符串无论如何都会返回str
,但不需要对其进行编码。 Unicode字符串代表文本,因此请保持这种方式:
title = metadata['xesam:title']
类似于artist
赋值,但获取可迭代的第一个元素:
artist = metadata['xesam:artist'][0]
接下来,在您的歌曲更新逻辑中,使用io.open
以UTF-8编码打开文件。这样可以直接写入Unicode字符串(文本),文件将处理编码。还可以使用with
语句在with
结束时自动关闭文件。
使用建议的更改进行编程:
import time
import dbus
import io
session_bus = dbus.SessionBus()
spotify_bus = session_bus.get_object("org.mpris.MediaPlayer2.spotify", "/org/mpris/MediaPlayer2")
spotify_properties = dbus.Interface(spotify_bus, "org.freedesktop.DBus.Properties")
def currentplaying():
metadata = spotify_properties.Get("org.mpris.MediaPlayer2.Player", "Metadata")
title = metadata['xesam:title']
artist = metadata['xesam:artist'][0]
playing = artist + " - " + title + " "
return playing
while True:
with io.open('currentsongspotify.txt', encoding='utf8') as filetxt:
oldtitle = filetxt.read()
newtitle = currentplaying()
if newtitle == oldtitle:
time.sleep(1)
else:
with io.open('currentsongspotify.txt','w',encoding='utf8') as filetxt: # save newtitle to file, overwriting existing data
filetxt.write(newtitle)
print 'new file saved:',newtitle