Python 3.5 YouTube API按关键字搜索:inf block除外

Question

我正在尝试使用YouTube API获取以下YouTube版本的代码。我还添加了我的个人API密钥和print的括号，但是代码不起作用。当我尝试在Pycharm中运行它时，我会得到：

except HttpError, e:
                    ^
SyntaxError: invalid syntax

我不明白为什么。这是代码：

#!/usr/bin/python
from apiclient.discovery import build
from apiclient.errors import HttpError
from oauth2client.tools import argparser

# Set DEVELOPER_KEY to the API key value from the APIs & auth > Registered apps
# tab of
#   https://cloud.google.com/console
# Please ensure that you have enabled the YouTube Data API for your project.
DEVELOPER_KEY = "MY_API_KEY"
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
def youtube_search(options):
  youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
    developerKey=DEVELOPER_KEY)
  # Call the search.list method to retrieve results matching the specified
  # query term.
  search_response = youtube.search().list(
    q=options.q,
    part="id,snippet",
    maxResults=options.max_results
  ).execute()
  videos = []
  channels = []
  playlists = []
  # Add each result to the appropriate list, and then display the lists of
  # matching videos, channels, and playlists.
  for search_result in search_response.get("items", []):
    if search_result["id"]["kind"] == "youtube#video":
      videos.append("%s (%s)" % (search_result["snippet"]["title"],
                                 search_result["id"]["videoId"]))
    elif search_result["id"]["kind"] == "youtube#channel":
      channels.append("%s (%s)" % (search_result["snippet"]["title"],
                                   search_result["id"]["channelId"]))
    elif search_result["id"]["kind"] == "youtube#playlist":
      playlists.append("%s (%s)" % (search_result["snippet"]["title"],
                                    search_result["id"]["playlistId"]))
  print ("Videos:n", "n".join(videos), "n")
  print ("Channels:n", "n".join(channels), "n")
  print ("Playlists:n", "n".join(playlists), "n")

if __name__ == "__main__":
  argparser.add_argument("--q", help="Search term", default="Rome")
  argparser.add_argument("--max-results", help="Max results", default=25)
  args = argparser.parse_args()
  try:
    youtube_search(args)
  except HttpError, e:
    print ("An HTTP error %d occurred:n%s" % (e.resp.status, e.content))

Answer 1

原始代码是为Python 2编写的，而当您将print转换为函数时（Python 3是必需的），您忘记了转换except块：

except HttpError as e:是python 3的例外方式，但您也可以使用except e:，除了所有异常外（可能不是您想要的）。

此外，您可以在此处阅读有关Python 2和3之间的更改的更多信息。有些人可能会让您感到惊讶！