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Get document DOCTYPE with BeautifulSoup

开发者 https://www.devze.com 2022-12-24 01:34 出处:网络
I\'ve just started tinkering with scrapy in conjunction with BeautifulSoup and I\'m wondering if I\'m missing something very obvious but I can\'t seem to figure out how to get the doctype of a returne

I've just started tinkering with scrapy in conjunction with BeautifulSoup and I'm wondering if I'm missing something very obvious but I can't seem to figure out how to get the doctype of a returned html document from the resulting soup object.

Given the f开发者_JAVA技巧ollowing html:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html lang="en"> 
<head> 
<meta charset=utf-8 />
<meta name="viewport" content="width=620" />
<title>HTML5 Demos and Examples</title> 
<link rel="stylesheet" href="/css/html5demos.css" type="text/css" /> 
<script src="js/h5utils.js"></script> 
</head> 
<body>
<p id="firstpara" align="center">This is paragraph <b>one</b>
<p id="secondpara" align="blah">This is paragraph <b>two</b>.
</html>

Can anyone tell me if there's a way of extracting the declared doctype from it using BeautifulSoup?


Beautiful Soup 4 has a class for DOCTYPE declarations, so you can use that to extract all the declarations at top level (though you're no doubt expecting one or none!)

def doctype(soup):
    items = [item for item in soup.contents if isinstance(item, bs4.Doctype)]
    return items[0] if items else None


You can go through top-level elements and check each to see whether it is a declaration. Then you can inspect it to find out what kind of declaration it is:

for child in soup.contents:
    if isinstance(child, BS.Declaration):
        declaration_type = child.string.split()[0]
        if declaration_type.upper() == 'DOCTYPE':
            declaration = child


You could just fetch the first item in soup contents:

>>> soup.contents[0]
u'DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"'
0

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