我在为网站的“下一页”网址获取XPath时遇到了麻烦。
HTML如下:
<div class="pagingcont">
<div class="right margintop" id="save_search_header_popup" style="width:550px;">
<div class="left marginleft" style="padding-top:1px;">
<div class="left save_search_env"><img src="/themes/LW1/refresh/images/envelope_icon.gif" alt="Save" /> </div>
<div class="left">
Save this search and receive email alerts of new listings
<input type="text" maxlength="100" value="Name this search" onfocus="doSavedSearchFocus(this,'Name this search');" style="width:120px;height:14px;color:Gray;"/>
</div>
</div>
<div class="left save_search_btn" style="margin-right:10px;"><img class="pointer" src="/themes/LW1/refresh/images/btn_save.gif" alt="Save" onclick="showPopup(document.getElementById('save_search_header_popup'), null, 'In order to be notified of new or updated properties, you need to be registered and signed in.');return false;"/></div>
</div>
<div class="left margintop marginleft" style="cursor:pointer;height:27px;" onclick="javascript:docompare(true);">
<div class="left"><img src="//www.landwatch.com/themes/LW1/images/comparebtn_btm.gif" style="margin-bottom:0px;"> </div>
<div class="left active" style="margin-top:4px;">COMPARE</div>
</div>
<div class="clear topline"></div>
<div class="clear margin">
<b>Page </b>
<span class="active" style="padding:3px 3px 3px 4px;border:solid 1px black;">1 </span> <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=2">2</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=3">3</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=4">4</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=5">5</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=6">6</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=7">7</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=8">8</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=9">9</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=10">10</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=11">11</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=12">12</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=13">13</a> | <a href="https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2c&pg=2">Next</a>
</div>
(我正在寻找的href是最右下角,这里看起来不方便......)
我的scrapy尝试以下方法:
next_page_url = response.xpath("//div[@class='pagingcont']//span//a[text()='Next']/href")
next_page_url = response.urljoin(next_page_url)
for href in response.css('div.propName a::attr(href)'):
url = response.urljoin(href.extract())
yield scrapy.Request(url, callback=self.parse_product_page)
yield scrapy.Request(next_page_url, callback=self.parse)
但每一次,scrapy都给了我第一页的结果,然后没有别的。所以我认为它没有有效地找到下一页。 next_page_url有什么问题?
答案 0 :(得分:2)
你的xpath有两个问题:
<span>
href
是一个属性,而不是一个节点,所以它应该是@href
。以下完整的工作示例。
from scrapy.spiders import Spider
from scrapy import Request
class LandSpider(Spider):
name = 'myspider'
start_urls = [
'https://www.landwatch.com/default.aspx?ct=r&type=5,37;268,6843&=&px=2000000&r.PSIZ=500%2C&pg=1']
def parse(self, response):
next_page_url = response.xpath(
"//div[@class='pagingcont']//a[text()='Next']/@href").extract_first()
for href in response.css('div.propName a::attr(href)'):
url = response.urljoin(href.extract())
yield Request(url, callback=self.parse_product_page)
yield Request(next_page_url, callback=self.parse)
def parse_product_page(self, response):
return response.xpath("//div[@class='detTitle']/text()").extract_first()
结果:
[
{"title": "Lulaton, Brantley County, Coast, GA Land For Sale - 936 Acres"},
{"title": "Oglethorpe County, GA Land For Sale - 515 Acres"},
{"title": "Dawsonville, Lumpkin County, GA Land For Sale - 525 Acres"},
{"title": "Wheeler County, GA Land For Sale - 594 Acres"},
{"title": "Cedartown, Polk County, GA Land For Sale - 1185.65 Acres"},
...
]
答案 1 :(得分:1)
首先,对于您展示的html示例,span
标记的父级没有a
,因此执行//span//a
并没有得到任何结果。所以也许你的xpath应该只是:
"//div[@class='pagingcont']//a[text()='Next']/href"
当然可能会更好。
现在你也没有得到你的python代码的值,这应该用.extract_first
完成,所以你的第一个next_page_url
变量(共享代码的第一行)是{{1实际上不是一个字符串。将其更改为:
Selector