如何过滤页面中的信息？

Question

我有这个代码：

import urllib
from bs4 import BeautifulSoup
import time

url = "http://www.downloadcrew.com/article/31121-magix_movie_edit_pro_2014_premium"
pageUrl = urllib.urlopen(url)
time.sleep(2)
soup = BeautifulSoup(pageUrl)

for a in  soup.select("div.downloadLink a[href]"):
    print "downloadlink: "+a["href"]

for b in soup.select("h1#articleTitle"):
    print b

for c in soup.select("table.detailsTable"):
    print c

我想要的是应用程序名称、更新日期、开发者和下载链接。 当我尝试运行它时，输出将是每个标签内的所有内容。

Answer 1

这里是得到你想要的代码：

import urllib
from bs4 import BeautifulSoup
import time

url = "http://www.downloadcrew.com/article/31121-magix_movie_edit_pro_2014_premium"
pageUrl = urllib.urlopen(url)
time.sleep(2)
soup = BeautifulSoup(pageUrl)

for a in  soup.select("div.downloadLink a[href]"):
    print "downloadlink: " + "?" + a["href"].split("?")[1].split(",")[0]

for b in soup.select("h1#articleTitle"):
    print b.contents[0].strip()

for c in soup.findAll("th"):
    if c.text == "Date Updated:":
        print c.parent.td.text
    elif c.text == "Developer:":
        print c.parent.td.text

但您无法使用该 URL 下载文件。您将需要检查 JavaScript 源文件以查看 javascript:checkDownload() 为获取实际文件位置所做的工作。