如何使用 xmlfeedspider 抓取 xml 提要 [英] How to scrape xml feed with xmlfeedspider

问题描述

我正在尝试使用以下格式抓取 xml 文件

I am trying to scrape an xml file with the below format

file_sample.xml:

<rss version="2.0">
 <channel>
   <item>
       <title>SENIOR BUDGET ANALYST (new)</title>
       <link>https://hr.example.org/psp/hrapp&SeqId=1</link>
       <pubDate>Wed, 18 Jul 2012 04:00:00 GMT</pubDate>
       <category>All Open Jobs</category>
   </item>
   <item>
       <title>BUDGET ANALYST (healthcare)</title>
       <link>https://hr.example.org/psp/hrapp&SeqId=2</link>
       <pubDate>Wed, 18 Jul 2012 04:00:00 GMT</pubDate>
       <category>All category</category>
   </item>
 </channel>
</rss>

下面是我的spider.py代码

class TestSpider(XMLFeedSpider):
    name = "testproject"
    allowed_domains = {"www.example.com"}
    start_urls = [
        "https://www.example.com/hrapp/rss/careers_jo_rss.xml"
        ]
    iterator = 'iternodes'
    itertag = 'channel'


    def parse_node(self, response, node):
        title = node.select('item/title/text()').extract()
        link  = node.select('item/link/text()').extract()
        pubdate  = node.select('item/pubDate/text()').extract()
        category  = node.select('item/category/text()').extract()
        item = TestprojectItem()
        item['title'] = title
        item['link'] = link
        item['pubdate'] = pubdate
        item['category'] = category
        return item

结果:

2012-07-25 13:24:14+0530 [testproject] DEBUG: Scraped from <200 https://hr.templehealth.org/hrapp/rss/careers_jo_rss.xml>
    {'title': [u'SENIOR BUDGET ANALYST (hospital/healthcare)',
               u'BUDGET ANALYST'],
     'link': [u'https://hr.example.org/psp/hrapp&SeqId=1',
               u'https://hr.example.org/psp/hrapp&SeqId=2'] 
     'pubdate': [u'Wed, 18 Jul 2012 04:00:00 GMT',
               u'Wed, 18 Jul 2012 04:00:00 GMT'] 
     'category': [u'All Open Jobs',
               u'All category'] 
      }

在这里你可以从上面的结果中观察到，来自相应标签的所有结果都被组合到一个列表中，但我想根据它们的单个项目标签进行映射，如下所示，因为我们为 html 抓取而这样做.

here as u can observe from the above result, all the results from the corresponding tags are combined in to single list, but i want to map according to their individual item tag like below as we do it for html scraping.

    {'title': u'SENIOR BUDGET ANALYST (hospital/healthcare)'
     'link': u'https://hr.example.org/psp/hrapp&SeqId=1'
     'pubdate': u'Wed, 18 Jul 2012 04:00:00 GMT'
     'category': u'All Open Jobs'
      }
    {'title': u'BUDGET ANALYST'
     'link': u'https://hr.example.org/psp/hrapp&SeqId=2' 
     'pubdate': u'Wed, 18 Jul 2012 04:00:00 GMT'
     'category': u'All category'
      }

我们如何根据单独的主标签(如上面的 item 标签)抓取 xml 标签数据.

How can we scrape xml tag data according to separate main tag like item tag above.

提前致谢.................

Thanks in advance.............

推荐答案

我推荐使用 feedparser:

feedparser.parse(url)

结果

{'bozo': 1,
 'bozo_exception': xml.sax._exceptions.SAXParseException("EntityRef: expecting ';'\n"),
 'encoding': u'utf-8',
 'entries': [{'link': u'https://hr.example.org/psp/hrapp&SeqId=1',
   'links': [{'href': u'https://hr.example.org/psp/hrapp&SeqId=1',
     'rel': u'alternate',
     'type': u'text/html'}],
   'tags': [{'label': None, 'scheme': None, 'term': u'All Open Jobs'}],
   'title': u'SENIOR BUDGET ANALYST (new)',
   'title_detail': {'base': u'',
    'language': None,
    'type': u'text/plain',
    'value': u'SENIOR BUDGET ANALYST (new)'},
   'updated': u'Wed, 18 Jul 2012 04:00:00 GMT',
   'updated_parsed': time.struct_time(tm_year=2012, tm_mon=7, tm_mday=18, tm_hour=4, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=200, tm_isdst=0)},
  {'link': u'https://hr.example.org/psp/hrapp&SeqId=2',
   'links': [{'href': u'https://hr.example.org/psp/hrapp&SeqId=2',
     'rel': u'alternate',
     'type': u'text/html'}],
   'tags': [{'label': None, 'scheme': None, 'term': u'All category'}],
   'title': u'BUDGET ANALYST (healthcare)',
   'title_detail': {'base': u'',
    'language': None,
    'type': u'text/plain',
    'value': u'BUDGET ANALYST (healthcare)'},
   'updated': u'Wed, 18 Jul 2012 04:00:00 GMT',
   'updated_parsed': time.struct_time(tm_year=2012, tm_mon=7, tm_mday=18, tm_hour=4, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=200, tm_isdst=0)}],
 'feed': {},
 'namespaces': {},
 'version': u'rss20'}

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