当该函数1在函数1中调用另一个函数2时，来自其他文件的函数1失败

Question

FileB.py 中的代码可以正常工作，但是当我从其他文件调用它时，有时会失败。我发现在下面的代码中调用函数“search_response”时它停止工作了。

文件A.py

from FileB import *
search = "stackoverflow"    
searchF(search)

文件B.py

from apiclient.discovery import build
from apiclient.errors import HttpError
from oauth2client.tools import argparser

search = "Google"    
def searchF(search):

  DEVELOPER_KEY = "REPLACE_ME"
  YOUTUBE_API_SERVICE_NAME = "youtube"
  YOUTUBE_API_VERSION = "v3"

打印“searchF started” - 有效

  def youtube_search(options):
    youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION,
      developerKey=DEVELOPER_KEY)

    search_response = youtube.search().list(
      q=options.q,
      type="video",
      part="id,snippet",
      maxResults=options.max_results
    ).execute()

打印“搜索响应已执行”不起作用

    search_videos = []

    for search_result in search_response.get("items", []):
      search_videos.append(search_result["id"]["videoId"])
    video_ids = ",".join(search_videos)

    video_response = youtube.videos().list(
      id=video_ids,
      part='snippet, contentDetails'
    ).execute()

    videos = []

    for video_result in video_response.get("items", []):
      videos.append("%s, (%s,%s)" % (video_result["snippet"]["title"],
                                video_result["contentDetails"],
                                video_result["contentDetails"]))
    find = "licensedContent': True"
    result = ', '.join(videos)
    print find in result

  if __name__ == "__main__":
    print "__main__"
    argparser.add_argument("--q", help="Search term", default=search)
    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

我将if __name__ == "__main__": 更改为if 1:，这有点用。但我认为这是一个可怕的解决方案。