2

尽管每个请求我有几个项目,但只有第一个(每个请求)正在进入管道,并且实际上保存为 Django 模型实例。

这是我的代码,我错过了什么?

# my_spider.py
class MySpider(CrawlSpider):
    name = 'my_spider'
    ...

    def parse(self, response):
        x = HtmlXPathSelector(response)
        item = MyDjangoItem()
        headings = x.select('//h2/text()').extract()
        for h in headings:
            item['name'] = h
            yield item

        url = 'http://example.com/next'  # I have custom rules for constructing (not extracting) next url
        yield Request(url, callback=self.parse)

# pipelines.py
class MyPipeline(object):
    def process_item(self, item, spider):
        if spider.name == 'my_spider':
            if item['name']:
                item.save()
        return item
4

1 回答 1

9

您需要MyDjangoItem在 for 循环内移动实例化,否则它总是会产生相同的对象。

# my_spider.py
class MySpider(CrawlSpider):
    name = 'my_spider'
    ...

    def parse(self, response):
        x = HtmlXPathSelector(response)

        headings = x.select('//h2/text()').extract()
        for h in headings:
            item = MyDjangoItem()
            item['name'] = h
            yield item

        url = 'http://example.com/next'  # I have custom rules for constructing (not extracting) next url
        yield Request(url, callback=self.parse)
于 2013-09-20T14:16:08.053 回答