Question

我试图从许多正在返回的网址中提取特定文本。我正在使用带有请求的Python 2.7和BeautifulSoup。

原因是我需要找到最新的URL，可以用最高的数字来识别＆＃34; DF_7＆＃34; 7是来自以下网址的最高。这个网址将被下载。请注意，每天都会添加新文件，这就是我需要检查编号最高的文件的原因。

一旦我在网址列表中找到最高编号，我就需要加入这个＆＃34; https://service.rl360.com/scripts/customer.cgi/SC/servicing/＆＃34;到编号最高的网址。最终产品应如下所示。 https://service.rl360.com/scripts/customer.cgi/SC/servicing/downloads.php?Reference=DF_7&SortField=ExpiryDays&SortOrder=Ascending

每次使用DF_递增时，网址看起来都是这样的

这是正确的做法吗？如果是这样我该怎么做呢。

由于

import base
import requests
import zipfile, StringIO, re
from lxml import html
from bs4 import BeautifulSoup

from base import os

from django.conf import settings

# Fill in your details here to be posted to the login form.
payload = {
    'USERNAME': 'xxxxxx',
    'PASSWORD': 'xxxxxx',
    'option': 'login'
}

headers = {'user-agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_5)     AppleWebKit/537.36 (KHTML, like Gecko) Chrome/59.0.3071.115 Safari/537.36'}

# Use 'with' to ensure the session context is closed after use.

with requests.Session() as s:
        p = s.post('https://service.rl360.com/scripts/customer.cgi?option=login', data=payload)

    # An authorised request.
    r = s.get('https://service.rl360.com/scripts/customer.cgi/SC/servicing/downloads.php?Folder=DataDownloads&SortField=ExpiryDays&SortOrder=Ascending', stream=True)
    content = r.text
    soup = BeautifulSoup(content, 'lxml')
    table = soup.find('table')
    links = table.find_all('a')
    print links

Answer 1

你可以直接到班级的最后一个链接＆＃34; tableid＆＃34;并按照以下方式打印它的href值：

href = soup.find_all("a", {'class':'tabletd'})[-1]['href']
base = "https://service.rl360.com/scripts/customer.cgi/SC/servicing/"
print (base + href)

从URL Python中提取特定文本

1 个答案: