Question

只有父项存在时才需要将父名称加入子行，否则不要将父名称加入行。

结果我需要有或没有父名称的子行。

我该如何实现？

SELECT t.*, cat.name AS cat_name
FROM products AS t
INNER JOIN category AS cat
ON category_id=cat.id
WHERE t.is_public!=2

Answer 1

只需将from scrapy.spider import BaseSpider from scrapy.selector import HtmlXPathSelector from scrapy.contrib.spiders import CrawlSpider, Rule from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor import scrapy from scrapy.contrib.linkextractors import LinkExtractor class CraigslistSampleItem(scrapy.Item): title = scrapy.Field() url = scrapy.Field() date = scrapy.Field() description=scrapy.Field() class SiteSpider(CrawlSpider): name = "newscrap" #download_delay = 2 allowed_domains = ['example.com'] start_urls = ['http://example.com/page/1'] items = {} def parse(self, response): sel = Selector(response) #requests =[] brands = sel.xpath("//div[@class='thumb']") for brand in brands: item = CraigslistSampleItem() url = brand.xpath("./a/@href")[0].extract() item['url'] = brand.xpath("./a/@href")[0].extract() item ["title"] = brand.xpath("./a/@title").extract() item ["date"] = brands.select("//span/text()").extract()[counter] counter=counter+1 request = Request(url,callback=self.parse_model, meta={'item':item}) yield request def parse_model(self, response): sel = Selector(response) models = sel.xpath("//*[@id='blocks-left']/div[1]/div/div[5]/p") for model in models: item = CraigslistSampleItem(response.meta["item"]) item ['description'] = model.xpath("//*[@id='blocks-left']/div[1]/div/div[5]/p")[0].extract() yield item替换为LEFT JOIN，所有产品都将包含在结果集中，var myNewFileObject = DriveApp.createFile("myXmlFile.xml", xmlContent, "text/xml");为INNER JOIN，用于没有父级/类别的产品。< / p>

仅在父项存在时才加入父名称

1 个答案: