Question

说我有以下HTML：

<p>
If everybody minded their own business, the world would go around a great deal faster than it does.
</p>

<p>
Who in the world am I? Ah, that's the great puzzle.
</p>

我希望能够找到包含我正在寻找的所有关键字的所有标签。例如。（例2和3不起作用）：

>>> len(soup.find_all(text="world"))
2

>>> len(soup.find_all(text="world puzzle"))
1

>>> len(soup.find_all(text="world puzzle book"))
0

我一直试图想出一个允许我搜索所有关键字的正则表达式，但似乎ANDing是不可能的（只有ORing）。

提前致谢！

Answer 1

像这样进行复杂匹配的最简单方法是write a function that performs the match，并将函数作为text参数的值传递。

def must_contain_all(*strings):                                                 
    def must_contain(markup):                                                   
        return markup is not None and all(s in markup for s in strings)         
    return must_contain

现在你可以得到匹配的字符串：

print soup.find_all(text=must_contain_all("world", "puzzle"))
# [u"\nWho in the world am I? Ah, that's the great puzzle.\n"]

要获取包含字符串的标记，请使用.parent运算符：

print [text.parent for text in soup.find_all(text=must_contain_all("world", "puzzle"))]
# [<p>Who in the world am I? Ah, that's the great puzzle.</p>]

Answer 2

您可能需要考虑使用lxml而不是BeautifulSoup。 lxml允许您通过XPath查找元素：

使用此锅炉板设置：

import lxml.html as LH
import re

html = """
<p>
If everybody minded their own business, the world would go around a great deal faster than it does.
</p>

<p>
Who in the world am I? Ah, that's the great puzzle.
</p>
"""

doc = LH.fromstring(html)

这会查找包含字符串<p>的所有world代码中的文字：

print(doc.xpath('//p[contains(text(),"world")]/text()'))
['\nIf everybody minded their own business, the world would go around a great deal faster than it does.\n', "\nWho in the world am I? Ah, that's the great puzzle.\n"]

这会找到包含<p>和world的所有puzzle代码中的所有文字：

print(doc.xpath('//p[contains(text(),"world") and contains(text(),"puzzle")]/text()'))
["\nWho in the world am I? Ah, that's the great puzzle.\n"]

Answer 3

这可能不是最有效的方法，但您可以尝试设置交叉点：

len(set(soup.find_all(text="world")
    & set(soup.find_all(text="book")
    & set(soup.find_all(text="puzzle")))

Answer 4

一点骨架（我使用的是lxml而不是BeautifulSoup，但你可以使用soup.findAll来适应它）：

html = """
<p>
If everybody minded their own business, the world would go around a great deal faster than it does.
</p>

<p>
Who in the world am I? Ah, that's the great puzzle.
</p>
"""

import lxml.html
import re

fragment = lxml.html.fromstring(html)
d = dict(
    (node, set(re.findall(r'\S+', node.text_content())))
    for node in fragment.xpath('//p'))

for node, it in d.iteritems():
    # then use set logic to go from here...

在BeautifulSoup中查找不同的字符串并返回包含标记

4 个答案: