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I need help with an lxml statement in python that extracts a meta-tag value

开发者 https://www.devze.com 2023-02-06 06:11 出处:网络
I need help fixing this lxml statement to extract the: http://www.etc../1tru.jpg link in the head section of

I need help fixing this lxml statement to extract the: http://www.etc../1tru.jpg link in the head section of http://www.yfrog.com/9d1truj

#This doe开发者_C百科sn't work!

# <link rel="image_src" href="http://img337.yfrog.com/img337/5023/1tru.jpg" />
def extract_imageurl(self, doc):
    try:
        self.url, = doc.xpath('//head//link[@rel="image_src"][1]/@href')
    except ValueError:
        self.url = "Error"

thanks


In [32]: doc.xpath('//head/link[@rel="image_src"]/@href')[0]
Out[32]: 'http://img337.yfrog.com/img337/5023/1tru.jpg'

Notice xpath returns a list of nodes:

In [25]: doc.xpath('//head/link')
Out[25]: [<Element link at 9c94c5c>, <Element link at 9c94b6c>]

Once you've specified [@rel="image_src"] there is only one node in the list. You can pick off the node with [0] after the xpath call.

In [29]: doc.xpath('//head/link[@rel="image_src"]')[0]
Out[29]: <Element link at 9c94c5c>

import lxml.html as lh
import urllib2

url=r'http://www.yfrog.com/9d1truj'
doc=lh.parse(urllib2.urlopen(url))
link=doc.xpath('//head/link[@rel="image_src"]/@href')[0]
print(link)
# http://img337.yfrog.com/img337/5023/1tru.jpg
0

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