Question

In Python I have a string containing the sourcecode of a website. Within this sourcecode I want to get the link within an tag, if the tag contains a specific substring.

The input e.g. looks like this:

AnyKindOfString <a href="http://www.link-to-get.com">SearchString</a> AndEvenMoreString

So what I want to tell Python is to search for SearchString in the all tags within string and give me the first found http://www.link-to-get.com back.

This should only work, if SearchString is within the tag - and it should also work, if "SearchString" is part (substring) of http://www.link-to-get.com.

I'm searching for an answer like more than 30 minutes know and the only thing I found for Python was simply to extract every (or only external or only internal) links from a string.

Anyone having an idea?

Thx in advance!

Answer 1

使用BeautifulSoup 3.2.1和python 2.7

from BeautifulSoup import BeautifulSoup

search_string = 'SearchString'

website_source = '<a href="http://www.link-to-get.com">SearchString</a> <a href="http://www.link-to-get.com">OtherString</a>\
                  <a href="http://www.link-to-getSearchString.com">otherString</a>'

soup = BeautifulSoup(website_source)

# this will return a list of lists that has the url's and the name for the link
anchors = [[row['href'], row.text] for row in soup.findAll('a') if row['href'].find(search_string) <> -1 or search_string in row.text]

# prints whole list
print anchors

#prints first list
print anchors[0]

# prints the url for the first list
print anchors[0][0]

问题似乎是我使用BeautifulSoup 3.2.1测试了上面的内容，它只适用于python 2.x，而你使用的是python 3.4，因此错误。
如果您安装BeautifulSoup4并尝试以下代码，它应该工作。另请注意，BeautifulSoup4适用于2.x和3.x.

请注意，以下内容尚未经过测试。

from bs4 import BeautifulSoup

search_string = 'SearchString'

website_source = '<a href="http://www.link-to-get.com">SearchString</a> <a href="http://www.link-to-get.com">OtherString</a>\
                  <a href="http://www.link-to-getSearchString.com">otherString</a>'

soup = BeautifulSoup(website_source)

# this will return a list of lists that has the url's and the name for the link
anchors = [[row['href'], row.text] for row in soup.findAll('a') if row['href'].find(search_string) != -1 or search_string in row.text]

# prints whole list
print(anchors)

# prints first list
print(anchors[0])

# prints the url for the first list
print(anchors[0][0])

Answer 2

我已经粗略化了一些应该有用的代码，至少它适用于你给出的示例字符串。

myString = 'AnyKindOfString <a href="http://www.link-to-get.com">SearchString</a> AndEvenMoreString'

theLimit = len(myString)
searchStringLinkPairs = []
tempStr = myString[:]
i =0


while i < theLimit:
    startLoc = tempStr.find('<a')
    endLoc = tempStr.find("</a")
    print startLoc,"\t",endLoc
    subStr = tempStr[startLoc:endLoc]
    startLink = subStr.find("\"")
    subTwo = subStr[startLink+1:]
    endLink = subTwo.find("\"")
    myLink = subStr[startLink+1:startLink+1+endLink]

    searchStringStart = subStr.find(">")
    searchString = subStr[searchStringStart+1:endLoc]

    if myLink != "" and searchString != "":
        searchStringLinkPairs.append([myLink, searchString])
    tempStr = tempStr[endLoc+1:]
    i = endLoc
    if startLoc == -1 or endLoc == -1:
        i = 10 * theLimit

print searchStringLinkPairs

Answer 3

可以在pyquery（http://pythonhosted.org/pyquery/index.html）+ lxml（http://lxml.de/tutorial.html）的帮助下完成，如下所示

from pyquery import PyQuery as pq
from lxml import etree

pq_obj = pq(etree.fromstring('<body><p>AnyKindOfString <a href="http://www.link-to-get.com">SearchString</a> AndEvenMoreString</p><p>this is another string goes here</p><a> other</a></body>'))
search_string = 'SearchString'

links = pq_obj('a')
for link in links:
    if search_string in link.text:
        attrib = link.attrib
        print attrib.get('href')

# output
# http://www.link-to-get.com

Python：在HTML <a> tag

3 个答案: