到目前为止,我的代码成功地将HTML代码从5个结果中提取出来,当给定subreddit的名称时。不管是要搜索的是单曲还是单曲。我想打开这个链接并将其发送到另一个类(imgurdl)
考虑到我目前的代码,最好的方法是什么?在
from bs4 import BeautifulSoup
import praw
from urllib2 import urlopen
import urllib2
import sys
from urlparse import urljoin
import config
import imgurdl
import requests
cache = []
soup = BeautifulSoup
def reddit_login():
r = praw.Reddit(username = USER,
password = config.password,
client_id = config.client_id,
client_secret = config.client_secret,
user_agent = " v0.3"
)
print("***********logged in successfully***********")
return r
def get_category_links(subredditName, r):
print("Grabbing subreddit...")
submissions = r.subreddit(subredditName).hot(limit=5)
print("Grabbing comments...")
#comments = subred.comments(limit = 200)
for submission in submissions:
htmlSource = requests.get(submission.url).text
print (htmlSource)
r = reddit_login()
get_category_links(sys.argv[1], r)
从循环中查看是否合适,然后从循环中获取它。这样就不需要通过html源代码了。在
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