Scrapy 不能同时使用 return 和 yield

问题描述：

这是我的代码

def parse(self, response):
    soup = BeautifulSoup(response.body)
    hxs = HtmlXPathSelector(response)
    sites = hxs.select('//div[@class="row"]')
    items = []

    for site in sites[:5]:
        item = TestItem()
        item['username'] = "test5"
        request =  Request("http://www.example.org/profile.php",  callback = self.parseUserProfile)
        request.meta['item'] = item
        **yield item**

    mylinks= soup.find_all("a", text="Next")
    if mylinks:
        nextlink = mylinks[0].get('href')
        yield Request(urljoin(response.url, nextlink), callback=self.parse)

def parseUserProfile(self, response):
    item = response.meta['item']
    item['image_urls'] = "test3"
    return item

现在我上面的工作，但我没有得到 item['image_urls'] = "test3"

Now my above works but with that i am not getting value of item['image_urls'] = "test3"

它是空的

现在如果使用 return request 而不是 yield item

Now if use return request instead of yield item

然后得到 cannot use return with generator

如果我删除这一行

yield Request(urljoin(response.url, nextlink), callback=self.parse)然后我的代码工作正常，我可以得到 image_urls 但我不能按照链接

yield Request(urljoin(response.url, nextlink), callback=self.parse) Then my code works fine and i can get image_urls but then i canot follow the links

那么有什么办法可以让我使用 return request 和 yield together 以便我得到 item_urls

So is there any way so that i can use return request and yield together so that i get the item_urls

答

看起来您遇到了机械错误.而不是:

Looks like you have a mechanical error. Instead of:

for site in sites[:5]:
    item = TestItem()
    item['username'] = "test5"
    request =  Request("http://www.example.org/profile.php",  callback = self.parseUserProfile)
    request.meta['item'] = item
    **yield item**

您需要:

for site in sites[:5]:
    item = TestItem()
    item['username'] = "test5"
    request =  Request("http://www.example.org/profile.php",  callback = self.parseUserProfile)
    request.meta['item'] = item
    yield request

Scrapy 不能同时使用 return 和 yield

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