Question

我在显示我想要的项目时遇到问题。我的代码如下：

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.http import request
from scrapy.selector import HtmlXPathSelector
from texashealth.items import TexashealthItem

class texashealthspider(CrawlSpider):

    name="texashealth"
    allowed_domains=['jobs.texashealth.org']
    start_urls=['http://jobs.texashealth.org/search/?&q=&title=Filter%3A%20title&facility=Filter%3A%20facility&location=Filter%3A%20city&date=Filter%3A%20date']

    rules=(
    Rule(SgmlLinkExtractor(allow=("search/",)), callback="parse_health", follow=True),
    #Rule(SgmlLinkExtractor(allow=("startrow=\d",)),callback="parse_health",follow=True),
    )

    def parse_health(self, response):
        hxs=HtmlXPathSelector(response)
    titles=hxs.select('//tbody/tr/td')
    items = []

    for titles in titles:
        item=TexashealthItem()
        item['title']=titles.select('span[@class="jobTitle"]/a/text()').extract()
        item['link']=titles.select('span[@class="jobTitle"]/a/@href').extract()
        item['shifttype']=titles.select('span[@class="jobShiftType"]/text()').extract()
        item['location']=titles.select('span[@class="jobLocation"]/text()').extract()
        items.append(item)
    print items
    return items

并且正在显示的输出在json格式中如下所示：

[
    TexashealthItem(location=[], link=[u'/job/Fort-Worth-ULTRASONOGRAPHER-II-Job-TX-76101/31553900/'], shifttype=[], title=[u'ULTRASONOGRAPHER II Job']), 
    TexashealthItem(location=[], link=[], shifttype=[u'Texas Health Fort Worth'], title=[]), 
    TexashealthItem(location=[u'Fort Worth, TX, US'], link=[], shifttype=[], title=[]), 
    TexashealthItem(location=[], link=[], shifttype=[], title=[]), 
    TexashealthItem(location=[], link=[u'/job/Kaufman-RN-Acute-ICU-Full-Time-Kaufman-Job-TX-75142/35466900/'], shifttype=[], title=[u'RN--Telemetry--Full Time--Kaufman Job']), 
    TexashealthItem(location=[], link=[], shifttype=[u'Texas Health Kaufman'], title=[]), 
    TexashealthItem(location=[u'Kaufman, TX, US'], link=[], shifttype=[], title=[]), 
    TexashealthItem(location=[], link=[], shifttype=[], title=[]), 
    TexashealthItem(location=[], link=[u'/job/Fort-Worth-NURSE-PRACTITIONER-Occ-Med-Full-Time-Alliance-Job-TX-76101/35465400/'], shifttype=[], title=[u'NURSE PRACTITIONER-Occ Med-Full Time-Alliance Job']), 
    TexashealthItem(location=[], link=[], shifttype=[u'Texas Health Alliance'], title=[]), 
    TexashealthItem(location=[u'Fort Worth, TX, US'], link=[], shifttype=[], title=[]), 
    TexashealthItem(location=[], link=[], shifttype=[], title=[])
]

如上所示，项目的参数以不同的间隔显示，即，它在一行中显示标题和链接，其余的输出在其他单独的行中显示。

我可以获得一个解决方案，以便我可以一次性显示所有参数吗？

感谢您的帮助

Answer 1

您应该循环表行 - tr元素，而不是表格单元格 - td元素。

我建议您使用hxs.select('//table[@id="searchresults"]/tbody/tr')，然后在每次循环迭代中使用.//span...

titles=hxs.select('//table[@id="searchresults"]/tbody/tr')
items = []
for titles in titles:
    item['title']=titles.select('.//span[@class="jobTitle"]/a/text()').extract()
    item['link']=titles.select('.//span[@class="jobTitle"]/a/@href').extract()
    item['shifttype']=titles.select('.//span[@class="jobShiftType"]/text()').extract()
    item['location']=titles.select('.//span[@class="jobLocation"]/text()').extract()
    items.append(item)
return items

Scrapy输出问题

1 个答案: