I need to extract the meta keywords from a web page using Python. I was thinking that this could be done using urllib or urllib2, but I'm not sure. Anyone have any ideas?
I am using Python 2.6 on Windows XP
I need to extract the meta keywords from a web page using Python. I was thinking that this could be done using urllib or urllib2, but I'm not sure. Anyone have any ideas?
I am using Python 2.6 on Windows XP
lxml is faster than BeautifulSoup (I think) and has much better functionality, while remaining relatively easy to use. Example:
52> from urllib import urlopen
53> from lxml import etree
54> f = urlopen( "http://www.google.com" ).read()
55> tree = etree.HTML( f )
61> m = tree.xpath( "//meta" )
62> for i in m:
..> print etree.tostring( i )
..>
<meta http-equiv="content-type" content="text/html; charset=ISO-8859-2"/>
Edit: another example.
75> f = urlopen( "http://www.w3schools.com/XPath/xpath_syntax.asp" ).read()
76> tree = etree.HTML( f )
85> tree.xpath( "//meta[@name='Keywords']" )[0].get("content")
85> "xml,tutorial,html,dhtml,css,xsl,xhtml,javascript,asp,ado,vbscript,dom,sql,colors,soap,php,authoring,programming,training,learning,b
eginner's guide,primer,lessons,school,howto,reference,examples,samples,source code,tags,demos,tips,links,FAQ,tag list,forms,frames,color table,w3c,cascading
style sheets,active server pages,dynamic html,internet,database,development,Web building,Webmaster,html guide"
BTW: XPath is worth knowing.
Another edit:
Alternatively, you can just use regexp:
87> f = urlopen( "http://www.w3schools.com/XPath/xpath_syntax.asp" ).read()
88> import re
101> re.search( "<meta name=\"Keywords\".*?content=\"([^\"]*)\"", f ).group( 1 )
101>"xml,tutorial,html,dhtml,css,xsl,xhtml,javascript,asp,ado,vbscript,dom,sql, ...etc...
...but I find it less readable and more error prone (but involves only standard module and still fits on one line).
<meta>
tag which 'name' attribute is 'Keywords' :) –
Sumbawa BeautifulSoup is a great way to parse HTML with Python.
Particularly, check out the findAll method: http://www.crummy.com/software/BeautifulSoup/documentation.html
Why not use a regular expression
keywordregex = re.compile('<meta\sname=
["\']keywords["\']\scontent=["\'](.*?)["\']\s/>')
keywordlist = keywordregex.findall(html)
if len(keywordlist) > 0:
keywordlist = keywordlist[0]
keywordlist = keywordlist.split(", ")
© 2022 - 2024 — McMap. All rights reserved.