当该函数1在函数函数中调用另一个函数1时，来自其他文件的function1失败1

Question

fileb.py中的代码工作正常，但是当我从另一个文件调用它时，在某个时刻失败。我发现在下面的代码中调用函数" search_response"时它停止工作。

filea.py

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

fileb.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"搜索启动" - works

  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"执行的search_response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:，这有点有效。但是我认为这是一个可怕的解决方案。