我的HTML代码包含大量类似结构的div ...下面是包含2个这样的div的代码摘录
<!-- 1st Div start -->
<div class="outer-container">
<div class="inner-container">
<a href="www.xxxxxx.com"></a>
<div class="abc xyz" title="verified"></div>
<div class="mody">
<div class="row">
<div class="col-md-5 col-xs-12">
<h2><a class="mheading primary h4" href="/c/my-llc"><strong>Top Dude, LLC</strong></a></h2>
<div class="mvsdfm casmhrn" itemprop="address">
<span itemprop="Address">1223 Industrial Blvd</span><br>
<span itemprop="Locality">Paris</span>, <span itemprop="Region">BA</span> <span itemprop="postalCode">123345</span>
</div>
<div class="hidden-device-xs" itemprop="phone" rel="mainPhone">
(800) 845-0000
</div>
</div>
</div>
</div>
</div>
</div>
<!-- 2nd Div start -->
<div class="outer-container">
<div class="inner-container">
<a href="www.yyyyyy.com"></a>
<div class="mody">
<div class="row">
<div class="col-md-5 col-xs-12">
<h2><a class="mheading primary h4" href="/c/my-llc"><strong>Fat Dude, LLC</strong></a></h2>
<div class="mvsdfm casmhrn" itemprop="address">
<span itemprop="Address">7890 Business St</span><br>
<span itemprop="Locality">Tokyo</span>, <span itemprop="Region">MA</span> <span itemprop="postalCode">987655</span>
</div>
<div class="hidden-device-xs" itemprop="phone" rel="mainPhone">
(800) 845-0000
</div>
</div>
</div>
</div>
</div>
</div>
所以这就是我想要Scrapy做的事情 ......
如果 class =＆＃34; div，外容器＆＃34; 包含另一个div title =＆＃34;已验证＆＃34; 就像在上面的第一个div中一样，它应该转到它上面的URL（即www.xxxxxx.com）并在该页面上获取其他一些字段。
如果没有div包含 title =＆＃34;已验证＆＃34; ，就像上面的第二个div一样，它应该获取下的所有数据div class =＆＃34; mody＆＃34; 。即公司名称（Fat Dude，LLC），地址，城市，州等......并且不遵循网址（即www.yyyyy.com）
那么如何在Scrapy搜寻器中应用此条件/逻辑。我在考虑使用BeautifulSoup，但不确定......
到目前为止我尝试了什么......
class MySpider(CrawlSpider):
name = 'dknfetch'
start_urls = ['http://www.xxxxxx.com/scrapy/all-listing']
allowed_domains = ['www.xxxxx.com']
def parse(self, response):
hxs = Selector(response)
soup = BeautifulSoup(response.body, 'lxml')
nf = NewsFields()
cName = soup.find_all("a", class_="mheading primary h4")
addrs = soup.find_all("span", itemprop_="Address")
loclity = soup.find_all("span", itemprop_="Locality")
region = soup.find_all("span", itemprop_="Region")
post = soup.find_all("span", itemprop_="postalCode")
nf['companyName'] = cName[0]['content']
nf['address'] = addrs[0]['content']
nf['locality'] = loclity[0]['content']
nf['state'] = region[0]['content']
nf['zipcode'] = post[0]['content']
yield nf
for url in hxs.xpath('//div[@class="inner-container"]/a/@href').extract():
yield Request(url, callback=self.parse)
当然，上面的代码会返回并抓取 div class =＆＃34; inner-container＆＃34; 下的所有网址，因为此代码中没有指定条件抓取，因为我不知道在哪里/如何设置它。
如果有人之前做过类似的事情，请分享。谢谢
答案 0 :(得分：1)
不需要使用BeautifulSoup，Scrapy自带选择器功能（也单独发布为parsel）。让我们使用您的HTML作一个例子：
html = u"""
<!-- 1st Div start -->
<div class="outer-container">
<div class="inner-container">
<a href="www.xxxxxx.com"></a>
<div class="abc xyz" title="verified"></div>
<div class="mody">
<div class="row">
<div class="col-md-5 col-xs-12">
<h2><a class="mheading primary h4" href="/c/my-llc"><strong>Top Dude, LLC</strong></a></h2>
<div class="mvsdfm casmhrn" itemprop="address">
<span itemprop="Address">1223 Industrial Blvd</span><br>
<span itemprop="Locality">Paris</span>, <span itemprop="Region">BA</span> <span itemprop="postalCode">123345</span>
</div>
<div class="hidden-device-xs" itemprop="phone" rel="mainPhone">
(800) 845-0000
</div>
</div>
</div>
</div>
</div>
</div>
<!-- 2nd Div start -->
<div class="outer-container">
<div class="inner-container">
<a href="www.yyyyyy.com"></a>
<div class="mody">
<div class="row">
<div class="col-md-5 col-xs-12">
<h2><a class="mheading primary h4" href="/c/my-llc"><strong>Fat Dude, LLC</strong></a></h2>
<div class="mvsdfm casmhrn" itemprop="address">
<span itemprop="Address">7890 Business St</span><br>
<span itemprop="Locality">Tokyo</span>, <span itemprop="Region">MA</span> <span itemprop="postalCode">987655</span>
</div>
<div class="hidden-device-xs" itemprop="phone" rel="mainPhone">
(800) 845-0000
</div>
</div>
</div>
</div>
</div>
</div>
"""
from parsel import Selector
sel = Selector(text=html)
for div in sel.css('.outer-container'):
if div.css('div[title="verified"]'):
url = div.css('a::attr(href)').extract_first()
print 'verified, follow this URL:', url
else:
nf = {}
nf['companyName'] = div.xpath('string(.//h2)').extract_first()
nf['address'] = div.css('span[itemprop="Address"]::text').extract_first()
nf['locality'] = div.css('span[itemprop="Locality"]::text').extract_first()
nf['state'] = div.css('span[itemprop="Region"]::text').extract_first()
nf['zipcode'] = div.css('span[itemprop="postalCode"]::text').extract_first()
print 'not verified, extracted item is:', nf
上一个代码段的结果是：
verified, follow this URL: www.xxxxxx.com
not verified, extracted item is: {'zipcode': u'987655', 'state': u'MA', 'address': u'7890 Business St', 'locality': u'Tokyo', 'companyName': u'Fat Dude, LLC'}
但是在Scrapy中你甚至不需要实例化Selector类，传递给回调的response对象中有一个快捷方式。此外，您不应该是CrawlSpider的子类，只需常规的Spider类即可。把它们放在一起：
from scrapy import Spider, Request
from myproject.items import NewsFields
class MySpider(Spider):
name = 'dknfetch'
start_urls = ['http://www.xxxxxx.com/scrapy/all-listing']
allowed_domains = ['www.xxxxx.com']
def parse(self, response):
for div in response.selector.css('.outer-container'):
if div.css('div[title="verified"]'):
url = div.css('a::attr(href)').extract_first()
yield Request(url)
else:
nf = NewsFields()
nf['companyName'] = div.xpath('string(.//h2)').extract_first()
nf['address'] = div.css('span[itemprop="Address"]::text').extract_first()
nf['locality'] = div.css('span[itemprop="Locality"]::text').extract_first()
nf['state'] = div.css('span[itemprop="Region"]::text').extract_first()
nf['zipcode'] = div.css('span[itemprop="postalCode"]::text').extract_first()
yield nf
我建议你熟悉Parsel的API：https://parsel.readthedocs.io/en/latest/usage.html
快乐刮！