Question

是否可能有一个“catch all”语句，该语句会影响for match in re.findall语句尚未添加的所有内容。

for submission in submissions:
    print (submission.url)
    for match in re.findall(".+imgur\.com+",submission.url):
        NonFilteredImgurImageList.append(submission.url)
    for match in re.findall("gfycat\.com",submission.url):
        GfycatUrlList.append(submission.url)
    else:
        NonStandardUrlList.append(submission.url)

我试过上面，但那没用。

有没有办法让它工作或更好的设置方式（我计划随着时间的推移添加更多的正则表达式搜索）。

Answer 1

看起来你不需要正则表达式。实际上，如果这些submission.url调用实际上包含多个结果，则会附加相应in个字符串的重复项。只需使用for submission in submissions: print(submission.url) if 'imgur.com' in submission.url: NonFilteredImgurImageList.append(submission.url) elif 'gifycat.com' in submission.url: GfycatUrlList.append(submission.url) else: NonStandardUrlList.append(submission.url)。

for submission in submissions:
    i = re.findall(expression1, submissions.url)
    g = re.findall(expression2, submissions.url)
    if i:
        results['imgur'] += i
    elif g:
        results['gfycat'] += g
    else:
        results['failures'].append(submissions.url)

如果此处实际需要正则表达式匹配，您将使用以下内容：

import getch
import threading

light_on = True
def countdown(count_from):
  global light_on
  print("Light will turn off in...")
  x = count_from
  while not light_on:
    print x
    time.sleep(1)
    x -= 1
  quit(0)

def testkey(keyname):
  global light_on
  while 1:
    char = getch.getch()
    if char == keyname:
      light_on = True

threading.Thread(target=testkey, args=('\n')).start()
while 1:
  if light_on: print("ON!")
  light_on = False
  threading.Thread(target=countdown, args=(10)).start()

在“for in”结束时抓住所有

1 个答案: