如何使用scrapy获取所有页面行元素

Question

我想废弃[关注http://lookup.mbon.org/verification/Search.aspx

首先我进入职业（护理）和姓氏（A *），然后获得列出的页面

我无法明智地获取列表页面，找到下一页代码的下面代码，

def on_page(self,response):
        forms = self.pagedata(response)
        pages = response.css("#datagrid_results").css("tr")[41].css("a::attr(href)").extract()
        page_continue=pages[-1]
        yield Request(url=response.url, callback=self.on_search,dont_filter=True,errback=self.handle_error)
        for page in pages:
            no_page=page[25:53]
            forms["__EVENTTARGET"] = no_page
            if page != page_continue:
                if(page != "javascript:__doPostBack('datagrid_results$ctl44$ctl00','')"):
                    yield FormRequest(url=response.url,callback=self.on_search, formdata = forms, dont_filter=True)

                elif page == "javascript:__doPostBack('datagrid_results$ctl44$ctl40','')" or "javascript:__doPostBack('datagrid_results$ctl44$ctl41','')" :
                    yield FormRequest(url=response.url,callback=self.on_search, formdata = forms, dont_filter=True) 

def on_search(self, response):
    print ("search started")
    pages = response.xpath("//table[@id='datagrid_results']/tr[42]/text()").extract()
    rows = response.xpath("//table[@id='datagrid_results']/tr/td[contains(@id, 'datagrid_results')]/a/@href").extract()
    for row in rows:
        url = response.urljoin(row) 
        yield Request(url = url, callback=self.on_detail_page)