<pre><code>import urllib
import re
import os
search = (raw_input('[!]Search: '))
site = "http://www.exploit-db.com/list.php?description="+search+"&author=&platform=&type=&port=&osvdb=&cve="
print site
source = urllib.urlopen(site).read()
founds = re.findall("href='/exploits/\d+",source)
print "\n[+]Search",len(founds),"Results\n"
if len(founds) >=1:
for found in founds:
found = found.replace("href='","")
print "http://www.exploit-db.com"+found
else:
print "\nCouldnt find anything with your search\n"
</code></pre>
<p>当我搜索漏洞时-数据库我的网站我只得出25个结果,我如何才能使它转到另一个页面或去通过25个结果。在</p>