通过lxml从根而不是元素开始进行xpath查找

问题描述：

我想在漂亮的汤中做同样的事情，找到 find_all 元素，并对其进行迭代以在每个迭代的元素中找到一些其他元素.即:

I want to do the same thing I do in beautiful soup, find_all elements and iterate through them to find some other_elements in each iterated elements. i.e.:

soup = bs4.BeautifulSoup(source)
articles = soup.find_all('div', class='v-card')
for article in articles:
    name = article.find('span', itemprop='name').text
    address = article.find('p', itemprop='address').text

现在，我尝试在lxml中执行相同的操作:

Now I try to do the same thing in lxml:

tree = html.fromstring(source)
items = tree.xpath('//div[@class="v-card"]')
for item in items:
    name = item.xpath('//span[@itemprop="name"]/text()')
    address = item.xpath('//p[@itemprop="address"]/text()')

...但是这会查找树中的所有匹配项，无论它们是否在当前 item 下.我该如何处理?

...but this finds all matches in the tree, regardless of whether they are under the current item. How can I approach this?

答

在后续查询中不要使用//作为前缀，该查询明确要求查询从根开始比你当前的元素.而是使用 .//进行相对查询:

Don't use // as prefix in the follow-up queries, which explicitly asks the query to start from the root rather than your current element. Instead, use .// for relative queries:

for item in tree.xpath('//div[@class="v-card"]'):
    name = item.xpath('.//span[@itemprop="name"]/text()'
    address = item.xpath('.//p[@itemprop="address"]/text()')

通过lxml从根而不是元素开始进行xpath查找

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