find_previous_siblings() 和 find_previous_sibling()find_previous_siblings( name , attrs , recursive , string , **kwargs ) find_previous_sibling( name , attrs , recursive , string , **kwargs ) 这2个方法通过 .previous_siblings 属性对当前tag的前面解析 [5] 的兄弟tag节点进行迭代, last_link = soup.find("a", id="link3")
last_link
# <a class="sister" href="http://example.com/tillie" id="link3">Tillie</a>
last_link.find_previous_siblings("a")
# [<a class="sister" href="http://example.com/lacie" id="link2">Lacie</a>,
# <a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>]
first_story_paragraph = soup.find("p", "story")
first_story_paragraph.find_previous_sibling("p")
# <p class="title"><b>The Dormouse's story</b></p>
find_all_next() 和 find_next()find_all_next( name , attrs , recursive , string , **kwargs ) find_next( name , attrs , recursive , string , **kwargs ) 这2个方法通过 .next_elements 属性对当前tag的之后的 [5] tag和字符串进行迭代, first_link = soup.a
first_link
# <a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>
first_link.find_all_next(string=True)
# [u'Elsie', u',\n', u'Lacie', u' and\n', u'Tillie',
# u';\nand they lived at the bottom of a well.', u'\n\n', u'...', u'\n']
first_link.find_next("p")
# <p class="story">...</p>
第一个例子中,字符串 “Elsie”也被显示出来,尽管它被包含在我们开始查找的<a>标签的里面.第二个例子中,最后一个<p>标签也被显示出来,尽管它与我们开始查找位置的<a>标签不属于同一部分.例子中,搜索的重点是要匹配过滤器的条件,并且在文档中出现的顺序而不是开始查找的元素的位置. find_all_previous() 和 find_previous()find_all_previous( name , attrs , recursive , string , **kwargs ) find_previous( name , attrs , recursive , string , **kwargs ) 这2个方法通过 .previous_elements 属性对当前节点前面 [5] 的tag和字符串进行迭代, first_link = soup.a
first_link
# <a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>
first_link.find_all_previous("p")
# [<p class="story">Once upon a time there were three little sisters; ...</p>,
# <p class="title"><b>The Dormouse's story</b></p>]
first_link.find_previous("title")
# <title>The Dormouse's story</title>
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