python 分解 HTML 以链接文本和目标
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Decomposing HTML to link text and target
提问by sundeep
Given an HTML link like
给定一个 HTML 链接，如
<a href="urltxt" class="someclass" close="true">texttxt</a>
how can I isolate the url and the text?
如何隔离 url 和文本？
Updates
更新
I'm using Beautiful Soup, and am unable to figure out how to do that.
我正在使用 Beautiful Soup，但我无法弄清楚如何做到这一点。
I did
我做了
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))
links = soup.findAll('a')
for link in links:
print "link content:", link.content," and attr:",link.attrs
i get
我明白了
*link content: None and attr: [(u'href', u'_redirectGeneric.asp?genericURL=/root /support.asp')]* ...
...
Why am i missing the content?
为什么我缺少内容？
edit: elaborated on 'stuck' as advised :)
编辑：按照建议详细说明“卡住”:)
回答by Harley Holcombe
Use Beautiful Soup. Doing it yourself is harder than it looks, you'll be better off using a tried and tested module.
使用美汤。自己动手比看起来更难，最好使用久经考验的模块。
EDIT:
编辑：
I think you want:
我想你想要：
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url).read())
By the way, it's a bad idea to try opening the URL there, as if it goes wrong it could get ugly.
顺便说一句，尝试在那里打开 URL 是一个坏主意，因为如果出错了它可能会变得丑陋。
EDIT 2:
编辑2：
This should show you all the links in a page:
这应该会显示页面中的所有链接：
import urlparse, urllib
from BeautifulSoup import BeautifulSoup
url = "http://www.example.com/index.html"
source = urllib.urlopen(url).read()
soup = BeautifulSoup(source)
for item in soup.fetchall('a'):
try:
link = urlparse.urlparse(item['href'].lower())
except:
# Not a valid link
pass
else:
print link
回答by Jerub
Here's a code example, showing getting the attributes and contents of the links:
这是一个代码示例，显示了获取链接的属性和内容：
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))
for link in soup.findAll('a'):
print link.attrs, link.contents
回答by Tom
Looks like you have two issues there:
看起来你有两个问题：
- link.contents, not link.content
- attrs is a dictionary, not a string. It holds key value pairs for each attribute in an HTML element. link.attrs['href'] will get you what you appear to be looking for, but you'd want to wrap that in a check in case you come across an a tag without an href attribute.
- link.content小号，不link.content
- attrs 是字典，而不是字符串。它保存 HTML 元素中每个属性的键值对。link.attrs['href'] 将为您提供您想要的内容，但您希望将其包装在支票中，以防您遇到没有 href 属性的 a 标签。
回答by nickf
Though I suppose the others mightbe correct in pointing you to using Beautiful Soup, they mightnot, and using an external library might be massively over-the-top for your purposes. Here's a regex which will do what you ask.
虽然我认为其他人在指出您使用 Beautiful Soup 方面可能是正确的，但他们可能不是，并且使用外部库可能会大大超出您的目的。这是一个正则表达式，可以满足您的要求。
/<a\s+[^>]*?href="([^"]*)".*?>(.*?)<\/a>/
Here's what it matches:
这是它匹配的内容：
'<a href="url" close="true">text</a>'
// Parts: "url", "text"
'<a href="url" close="true">text<span>something</span></a>'
// Parts: "url", "text<span>something</span>"
If you wanted to get justthe text (eg: "textsomething" in the second example above), I'd just run another regex over it to strip anything between pointed brackets.
如果你希望得到公正的文本（例如：在上面的第二个例子“textsomething”），我只是运行另一个正则表达式在它剥离尖括号之间的任何东西。
