Question

我的代码......

foo = fromstring(my_html)

它提出了这个警告......

UserWarning: No parser was explicitly specified, so I'm using the best available HTML parser for this system ("html.parser"). This usually isn't a problem, but if you run this code on another system, or in a different virtual environment, it may use a different parser and behave differently.

To get rid of this warning, change this:

 BeautifulSoup([your markup])

to this:

 BeautifulSoup([your markup], "html.parser")

  markup_type=markup_type))

我已经尝试将字符串'html.parser'传递给它但是这不起作用，因为它给我一个错误，说明字符串不是可调用的，所以我尝试了html.parser然后我查看了lxml模块看看我是否能找到另一个解析器，但却无法找到。我查看了python stdlib并看到在2.7中有一个名为HTMLParser，所以我导入了它并输入了beautifulsoup=HTMLParser，但这也没有用。

我应该传递给fromstring的可调用者在哪里？

EDIT添加了尝试的解决方案：

from lxml.html.soupparser import fromstring
wiktionary_page = fromstring(wiktionary_page.read(), features="html.parser" )

和这个

from lxml.html.soupparser import BeautifulSoup
wiktionary_page = fromstring(wiktionary_page.read(), beautifulsoup=lambda s: BeautifulSoup(s, "html.parser"))

Answer 1

您可以传递将设置解析器的 features 关键字。

tree = lxml.html.soupparser.fromstring("<p>foo</p>", features="html.parser" )

fromstring 中发生的事情 _parser 会被调用，但我认为行中有一个错误 bsargs ['features'] = ['html.parser '] ，它应该是bsargs['features'] = 'html.parser'：

def _parse(source, beautifulsoup, makeelement, **bsargs):
    if beautifulsoup is None:
        beautifulsoup = BeautifulSoup
    if hasattr(beautifulsoup, "HTML_ENTITIES"):  # bs3
        if 'convertEntities' not in bsargs:
            bsargs['convertEntities'] = 'html'
    if hasattr(beautifulsoup, "DEFAULT_BUILDER_FEATURES"):  # bs4
        if 'features' not in bsargs:
            bsargs['features'] = ['html.parser']  # use Python html parser
    tree = beautifulsoup(source, **bsargs)
    root = _convert_tree(tree, makeelement)
    # from ET: wrap the document in a html root element, if necessary
    if len(root) == 1 and root[0].tag == "html":
        return root[0]
    root.tag = "html"
    return root

你也可以使用lambda：

from lxml.html.soupparser import BeautifulSoup
import lxml.html.soupparser

tree = lxml.html.soupparser.fromstring("<p>foo</p>", beautifulsoup=lambda s: BeautifulSoup(s, "html.parser"))

两者都禁止任何警告：

In [13]: from lxml.html import soupparser

In [14]: tree = soupparser.fromstring("<p>foo</p>", features="html.parser" )
In [15]: from lxml.html.soupparser import BeautifulSoup

In [16]: import lxml.html.soupparser


In [17]: tree = lxml.html.soupparser.fromstring("<p>foo</p>", beautifulsoup=lambda s: BeautifulSoup(s, "html.parser"))

python lxml.html.soupparser.fromstring引发恼人的警告

1 个答案: