how could I get all email address in a html page? Have any modle can do
this?
like the htmldata.urlextract .
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for example:
re.sub(']+)+\s?>[^<^>]*','',' asd gahttp://www.sine.com"; class="wordstyle"> asdgasdghae rha')
I wish to get the return value "asd ga asdgasdghae rha",how do do?
I have a impression on "%" and "{number}",but forgot how to use them.
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to Diez B. Roggisch :
no, I won't to use this way, its efficiency is very low as I need to
replace all the 'http://mail.python.org/mailman/listinfo/python-list
is there any lib to help to clear the absolute url in a html file? for
example, 'http://www.sina.com/' should be clear ,but "/image/asd.gif"
should be reserved.
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oh~~~! offer my thanks to Tim Roberts and all persons above!
I see now, it's the different url causes!
contents can only be got from the later (real ) url.
I made a mistick not to look at the different urls taking effect.
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yeah,u r right, the page uses chinese.(I'm a chinese too.^_^,)
using urllib2.urlopen('').read(),I can't get the contents
between '' and '' ,the reason isn't the chinese encoding
but the 'no-cache' set,I think.
I want to get the contents between
can you find the problem why i can'
using web browser can get page's content formally, but when use
urllib2.open("http://tech.163.com/2004w11/12732/2004w11_1100059465339.html";).read()
the result is
http://tech.163.com/04/1110/12/14QUR2BR0009159H.html";>
?y?ú'ò?aò3??...
,I think the reson is the no-cache, are there person would he
thanks a lot.
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but I want to ask: how to find out explaination about the parameters of
'open()' function?
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thank james;
and I have got a way to resolve it :
urllib.urlretrieve('http://x.x.x.x/library.zip','library.zip')
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I wrote as following:
content=urllib2.urlopen("http://www.sdfagfa.org/asasg/a.zip";).read()
f=open("d:\\a.zip",'a+',1)
f.writelines(content)
the zip file has been download to my machine,but when be unzip , error
occurs:
" there is an error at the end of the file"
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