google youtube api on Django
I'm working on a website to get videos via key words using google api on Django. The function for calling api works well individually, but once I called the function in view.py and run the server, it returns server error. why?
def youtube_search(options):
DEVELOPER_KEY = "AIzaSyBQG9ozouzPofCHE4J-BHdUeSjqqtemnc0"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
developerKey=DEVELOPER_KEY)
parser = argparse.ArgumentParser()
parser.add_argument("--q", help="Search term", default=options)
parser.add_argument("--max-results", help="Max results", default=1) #
the number of default is the number of output videos
options= parser.parse_args()
search_response = youtube.search().list(
q=options.q,
part="id,snippet",
maxResults=options.max_results
).execute()
videos =
videoid =
channels =
playlists =
result =
for search_result in search_response.get("items", ):
if search_result["id"]["kind"] == "youtube#video":
videos.append( search_result["snippet"]["title"])
videoid.append( search_result["id"]["videoId"])
for line in videos:
result.append(line)
for line1 in videoid:
line1 = "http://www.youtube.com/watch?v="+line1
result.append(line1)
return result
and here are the view.py calling the function:
def index2(request):
if request.method == "POST":
search=request.POST.get("search:",None)
class1 = predict.predict(search)
#list =[search, 'other corresponding tags']
api = YTapi.youtube_search(search)
video = api[0]
videoid = api[1]
return render(request, "index2.html","video":video,"videoid":videoid)
django
add a comment |
I'm working on a website to get videos via key words using google api on Django. The function for calling api works well individually, but once I called the function in view.py and run the server, it returns server error. why?
def youtube_search(options):
DEVELOPER_KEY = "AIzaSyBQG9ozouzPofCHE4J-BHdUeSjqqtemnc0"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
developerKey=DEVELOPER_KEY)
parser = argparse.ArgumentParser()
parser.add_argument("--q", help="Search term", default=options)
parser.add_argument("--max-results", help="Max results", default=1) #
the number of default is the number of output videos
options= parser.parse_args()
search_response = youtube.search().list(
q=options.q,
part="id,snippet",
maxResults=options.max_results
).execute()
videos =
videoid =
channels =
playlists =
result =
for search_result in search_response.get("items", ):
if search_result["id"]["kind"] == "youtube#video":
videos.append( search_result["snippet"]["title"])
videoid.append( search_result["id"]["videoId"])
for line in videos:
result.append(line)
for line1 in videoid:
line1 = "http://www.youtube.com/watch?v="+line1
result.append(line1)
return result
and here are the view.py calling the function:
def index2(request):
if request.method == "POST":
search=request.POST.get("search:",None)
class1 = predict.predict(search)
#list =[search, 'other corresponding tags']
api = YTapi.youtube_search(search)
video = api[0]
videoid = api[1]
return render(request, "index2.html","video":video,"videoid":videoid)
django
Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set theDEBUG = Trueon django'ssettings.pyto get a detail error message.
– Himal
Nov 10 at 3:43
add a comment |
I'm working on a website to get videos via key words using google api on Django. The function for calling api works well individually, but once I called the function in view.py and run the server, it returns server error. why?
def youtube_search(options):
DEVELOPER_KEY = "AIzaSyBQG9ozouzPofCHE4J-BHdUeSjqqtemnc0"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
developerKey=DEVELOPER_KEY)
parser = argparse.ArgumentParser()
parser.add_argument("--q", help="Search term", default=options)
parser.add_argument("--max-results", help="Max results", default=1) #
the number of default is the number of output videos
options= parser.parse_args()
search_response = youtube.search().list(
q=options.q,
part="id,snippet",
maxResults=options.max_results
).execute()
videos =
videoid =
channels =
playlists =
result =
for search_result in search_response.get("items", ):
if search_result["id"]["kind"] == "youtube#video":
videos.append( search_result["snippet"]["title"])
videoid.append( search_result["id"]["videoId"])
for line in videos:
result.append(line)
for line1 in videoid:
line1 = "http://www.youtube.com/watch?v="+line1
result.append(line1)
return result
and here are the view.py calling the function:
def index2(request):
if request.method == "POST":
search=request.POST.get("search:",None)
class1 = predict.predict(search)
#list =[search, 'other corresponding tags']
api = YTapi.youtube_search(search)
video = api[0]
videoid = api[1]
return render(request, "index2.html","video":video,"videoid":videoid)
django
I'm working on a website to get videos via key words using google api on Django. The function for calling api works well individually, but once I called the function in view.py and run the server, it returns server error. why?
def youtube_search(options):
DEVELOPER_KEY = "AIzaSyBQG9ozouzPofCHE4J-BHdUeSjqqtemnc0"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
developerKey=DEVELOPER_KEY)
parser = argparse.ArgumentParser()
parser.add_argument("--q", help="Search term", default=options)
parser.add_argument("--max-results", help="Max results", default=1) #
the number of default is the number of output videos
options= parser.parse_args()
search_response = youtube.search().list(
q=options.q,
part="id,snippet",
maxResults=options.max_results
).execute()
videos =
videoid =
channels =
playlists =
result =
for search_result in search_response.get("items", ):
if search_result["id"]["kind"] == "youtube#video":
videos.append( search_result["snippet"]["title"])
videoid.append( search_result["id"]["videoId"])
for line in videos:
result.append(line)
for line1 in videoid:
line1 = "http://www.youtube.com/watch?v="+line1
result.append(line1)
return result
and here are the view.py calling the function:
def index2(request):
if request.method == "POST":
search=request.POST.get("search:",None)
class1 = predict.predict(search)
#list =[search, 'other corresponding tags']
api = YTapi.youtube_search(search)
video = api[0]
videoid = api[1]
return render(request, "index2.html","video":video,"videoid":videoid)
django
django
asked Nov 10 at 2:54
Zhengyang
1
1
Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set theDEBUG = Trueon django'ssettings.pyto get a detail error message.
– Himal
Nov 10 at 3:43
add a comment |
Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set theDEBUG = Trueon django'ssettings.pyto get a detail error message.
– Himal
Nov 10 at 3:43
Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set the
DEBUG = True on django's settings.py to get a detail error message.– Himal
Nov 10 at 3:43
Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set the
DEBUG = True on django's settings.py to get a detail error message.– Himal
Nov 10 at 3:43
add a comment |
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Hello, Welcome to SO. You might wanna update the question with the error message you are getting. Might have set the
DEBUG = Trueon django'ssettings.pyto get a detail error message.– Himal
Nov 10 at 3:43