Python中从subreddits中获取Imgur链接

2024-04-24 13:17:53 发布

您现在位置:Python中文网/ 问答频道 /正文

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) 

Tags: 代码fromimportclientconfiggetsysurllib2
1条回答
网友
1楼 · 发布于 2024-04-24 13:17:53

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

 for submission in submissions:
    link = submission.url
    if "imgur.com/a/" in link:
        #Send to imgur album downloader
    elif link.endswith(".jpg") or link.endswith(".png"):
        #Sent to image downloader
    elif "imgur.com/" in link:
        #Send to single image imgur downloader

相关问题 更多 >