獲得元素我有一個XML像下面下面的XPath從XML
<li class="expandSubItem">
<span class="expandSubLink">Popular Neighborhoods</span>
<ul class="secondSubNav" style="top:-0.125em;">
<li class="subItem">
<a class="subLink" href="/Hotels-g187147-zfn7236765-Paris_Ile_de_France-Hotels.html">Quartier Latin Hotels</a>
</li>
</ul>
</li>
<li class="expandSubItem">
<span class="expandSubLink">Popular Paris Categories</span>
<ul class="secondSubNav" style="top:-0.125em;">
<li class="subItem">
<a class="subLink" href="/HotelsList-Paris-Cheap-Hotels-zfp10420.html">Paris Cheap Hotels</a>
</li>
</ul>
</li>
我想在「熱門巴黎類別」的所有鏈接。我使用了類似//li//a/@href/following::span[text()='Popular Singapore Categories']
的東西,但沒有得到任何結果。任何想法如何得到正確的結果?這是我寫的python代碼片段。
t_url = 'https://www.tripadvisor.com/Tourism-g187147-Paris_Ile_de_France-Vacations.html'
page = requests.get(t_url, timeout=30)
tree = html.fromstring(page.content)
links = tree.xpath('//li[span="Popular Paris Categories"]//a/@href')
print links
'// li [span ='熱門巴黎分類']/ul/li/a/@ href' – har07
它沒有用,因爲「span」和「ul」在xpath的同一層。 –
是的,'span'和'ul'處於同一水平,並不重要。查看演示(或者在您喜歡的任何XPath測試器中嘗試它):http://xpatheval.apphb.com/3849byFx2 – har07