无法获得最简单的管道示例在scrapy中工作

Question

这是我的简单代码,我没有得到它的工作.

我是子类 initspider

这是我的代码

class MytestSpider(InitSpider):
    name = 'mytest'
    allowed_domains = ['example.com']
    login_page = 'http://www.example.com'
    start_urls = ["http://www.example.com/ist.php"]

    def init_request(self):
        #"""This function is called before crawling starts."""
        return Request(url=self.login_page, callback=self.parse)

    def parse(self, response):
        item = MyItem()
        item['username'] = "mytest"
        return item

管道

class TestPipeline(object):
    def process_item(self, item, spider):
            print item['username']

如果尝试打印该项目,我会发出相同的错误

我得到的错误是

 File "crawler/pipelines.py", line 35, in process_item
            myitem.username = item['username']
        exceptions.TypeError: 'NoneType' object has no attribute '__getitem__'

我的问题是InitSpider.我的pieplines没有得到项目对象

items.py

class MyItem(Item):
    username = Field()

setting.py

BOT_NAME = 'crawler'

SPIDER_MODULES = ['spiders']
NEWSPIDER_MODULE = 'spiders'


DOWNLOADER_MIDDLEWARES = {

    'scrapy.contrib.downloadermiddleware.cookies.CookiesMiddleware': 700 # <-
}

COOKIES_ENABLED = True
COOKIES_DEBUG = True


ITEM_PIPELINES = [

'pipelines.TestPipeline',


]

IMAGES_STORE = '/var/www/htmlimages'

Answer 1

pipelines.TestPipeline缺少订单号。它应该是类似的东西ITEM_PIPELINES = {'pipelines.TestPipeline': 900}。