Home/ Questions/Q 9264527
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-18T13:50:57+00:00

I have a page with next code: <HTML> <HEAD> <TITLE>smth</TITLE> <META HTTP-EQUIV=Content-Type CONTENT=text/html; charset=UTF-8>

  • 0

I have a page with next code:

<HTML>
<HEAD>
<TITLE>smth</TITLE>
<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=UTF-8">
</HEAD>
<BODY>
<div id="doc" class="searchN">
<div id="hd" style="border-bottom:0;">
    <a id="logo" class="logoN" href="/" alt="logo" title="open project"></a>

</div> 
    <div id="bd-cross">    
        <ol class="site" start=1>

            <li class="">
                <a href="url/">Smth</a>
                <div class="ref">
                <a href="News_and_Media/">Regional: Europe:</a>
          </div>    
            </li>

            <li class="">
                <a href="url2">Descr3</a> 
                <div class="ref">
                <a href="url3">Descr3</a>   
          </div>    
            </li>
....
</BODY>
</HTML>

I need to check <li class=""> tag presence at page. I use Python+RegExp:

import re
import urllib2
url = 'url'
#Parse it
MainPage = urllib2.urlopen(url).read()
Li = re.findall("<div id=\"bd-cross\">*<li class=\"\">*</li>", MainPage)
try:
    if Li:
        print "Li tag on " +url+ ": Yes"
    else:
        print "Li tag on " +url+ ": No"
except:
    print "Error"

output is No but it should be Yes ’cause page contains it tag. If I print Li it output ‘[ ]’.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You should use a package such as BeautifulSoup or lxml.html.soupparser, it will make your life much much easier. With the latter, you could do what you want as follows:

    >>> import lxml.html.soupparser
>>> MainPage = urllib2.urlopen(url).read()
>>> HtmlDoc = lxml.html.soupparser.fromstring(MainPage)
>>> Elements = HmtlDoc.xpath('//div[@id="bd-cross"]//li[@class=""]')
>>> if len(Elements) > 0:
>>>     print 'Yes'
>>> else:
>>>     print 'No'
    • 0