我想创建一个允许用户将视频上传到Youtube的应用程序。 这是我尝试过的代码(改编自http://code.google.com/p/gdata-python-client/source/browse/tests/gdata_tests/youtube/service_test.py):
my_media_group = gdata.media.Group(
title = gdata.media.Title(text=title),
description = gdata.media.Description(description_type='plain',
text=description),
keywords = gdata.media.Keywords(text= tags),
category = gdata.media.Category(
text='Autos',
scheme='http://gdata.youtube.com/schemas/2007/categories.cat',
label='Autos'),
player=None
)
# Set Geo location to 37,-122 lat, long
where = gdata.geo.Where()
where.set_location((37.0,-122.0))
video_entry = gdata.youtube.YouTubeVideoEntry(media=my_media_group, geo=where)
print "Creating a video entry: done."
print "Uploading video"
new_entry = self.client.InsertVideoEntry(video_entry, filepath)
print "Done uploading video."
它总是停留在倒数第二行:
new_entry = self.client.InsertVideoEntry(video_entry, filepath)
那有什么不对?是否有关于filepath的要求(例如,路径'C:\ video.avi'是否有效?)
事实上,我只需要一种上传方式,所以请提出任何可能的解决方案。
编辑:1。我尝试嵌入Youtube上传小部件,但它似乎不起作用:(只有一个选项可以从网络摄像头加载,但这也不起作用。)
<iframe id="widget" type="text/html" width="640" height="390"
src="https://www.youtube.com/upload_embed" frameborder="0"></iframe>
编辑2:
我将代码更改为:
def getYtSession(self, email = '', password =''):
# if we do not want to reuse our session.
if email != '' and password != '':
yt_service = gdata.youtube.service.YouTubeService()
yt_service.email = email
yt_service.password = password
#login.
try:
print yt_service.ProgrammaticLogin()
self.email = email
self.password = password
self.logged_in = True
self.emit(QtCore.SIGNAL('doneLogin(QString)'), QtCore.QString(email + " " + password))
return yt_service
except:
#Display a warning dialog.
self.logged_in = False
self.emit(QtCore.SIGNAL('failedLogin()'))
return None
elif self.yt_service: # we want to reuse the session.
return self.yt_service
else:
return gdata.youtube.service.YouTubeService()
并在上传功能中:
self.yt_service = self.getYtSession(self.email, self.password)
try:
new_entry = self.yt_service.InsertVideoEntry(video_entry, filepath)
print "Done uploading video."
self.emit(QtCore.SIGNAL('doneUpload(QString)'), QtCore.QString('Title: ' + title + '\nPath: ' + filepath))
except:
print "Stack trace:"
traceback.print_stack()
它仍然不起作用。还有一个问题:堆栈跟踪没有任何帮助:
Stack trace:Login successfully.
文件“D:\ workplace \ simple-media-player \ trunk \ MyPlayer \ src \ QtThread.py”,第19行,运行中 self.function(* self.args,** self.kwargs) 在doRealUpload中输入文件“D:\ workplace \ simple-media-player \ trunk \ MyPlayer \ src \ videoplayer.py”,第490行 traceback.print_stack()
该消息(“登录成功。”)由发出信号doneLogin(QString)时调用的函数打印。如果我不放
new_entry = self.yt_service.InsertVideoEntry(video_entry, filepath)
print "Done uploading video."
self.emit(QtCore.SIGNAL('doneUpload(QString)'), QtCore.QString('Title: ' + title + '\nPath: ' + filepath))
在try-except中,然后它什么都不打印。
更多信息:我将视频信息设置如下: 路径:C:/Users/huynh/Desktop/REcord1.wmv 名称:'测试视频' 标签:'test,youtube' 描述:fafafa
答案 0 :(得分:0)