如何从管道中执行特定的蜘蛛，而无需再次激活它

Question

Introduction

The website I'm scrapping has two urls:

• /top lists top players
• /player/{name} shows player with name {name} info

From the first URL, I get the player name and position then I'm able to call the second URL using the given name. My current goal is to store all the data on a database.

The problem

I created two spiders. The first, which crawls /top and the second which crawls /player/{name} for each player the first spider has found. However, to be able to insert the first spider data into the database, I need to call the profile spider because it is a foreign key, as noted on the following queries:

INSERT INTO top_players (player_id, position) values (1, 1)

INSERT INTO players (name) values ('John Doe')

Question

Is it possible to execute a spider from the Pipeline just to get the spider results? I mean, the called spider should not activate the pipeline again.

Answer 1

i would suggest you to have more control over the scraping process. Especially with grabbing the name,position from the first page and detail page. try this:

# -*- coding: utf-8 -*-
import scrapy

class MyItem(scrapy.Item):
    name = scrapy.Field()
    position= scrapy.Field()
    detail=scrapy.Field() 
class MySpider(scrapy.Spider):

    name = '<name of spider>'
    allowed_domains = ['mywebsite.org']
    start_urls = ['http://mywebsite.org/<path to the page>']

    def parse(self, response):

        rows = response.xpath('//a[contains(@href,"<div id or class>")]')

        #loop over all links to stories
        for row in rows:
            myItem = MyItem() # Create a new item
            myItem['name'] = row.xpath('./text()').extract() # assign name from link
            myItem['position']=row.xpath('./text()').extract() # assign position from link
            detail_url = response.urljoin(row.xpath('./@href').extract()[0]) # extract url from link
            request = scrapy.Request(url = detail_url, callback = self.parse_detail) # create request for detail page with story
            request.meta['myItem'] = myItem # pass the item with the request
            yield request

    def parse_detail(self, response):
        myItem = response.meta['myItem'] # extract the item (with the name) from the response
        text_raw = response.xpath('//font[@size=3]//text()').extract() # extract the detail (text)
        myItem['detail'] = ' '.join(map(unicode.strip, text_raw)) # clean up the text and assign to item
        yield myItem # return the item