python - 如何从外部 python 脚本正确运行爬虫并获取其项目输出
问题描述
所以我正在制作几个刮板,现在我正在尝试制作一个脚本,该脚本运行具有从数据库收集的 URL 的相应蜘蛛,但我找不到这样做的方法。
我的蜘蛛中有这个:
class ElCorteIngles(scrapy.Spider):
name = 'ElCorteIngles'
url = ''
DEBUG = False
def start_requests(self):
if self.url != '':
yield scrapy.Request(url=self.url, callback=self.parse)
def parse(self, response):
# Get product name
try:
self.p_name = response.xpath('//*[@id="product-info"]/h2[1]/a/text()').get()
except:
print(f'{CERROR} Problem while getting product name from website - {self.name}')
# Get product price
try:
self.price_no_cent = response.xpath('//*[@id="price-container"]/div/span[2]/text()').get()
self.cent = response.xpath('//*[@id="price-container"]/div/span[2]/span[1]/text()').get()
self.currency = response.xpath('//*[@id="price-container"]/div/span[2]/span[2]/text()').get()
if self.currency == None:
self.currency = response.xpath('//*[@id="price-container"]/div/span[2]/span[1]/text()').get()
self.cent = None
except:
print(f'{CERROR} Problem while getting product price from website - {self.name}')
# Join self.price_no_cent with self.cent
try:
if self.cent != None:
self.price = str(self.price_no_cent) + str(self.cent)
self.price = self.price.replace(',', '.')
else:
self.price = self.price_no_cent
except:
print(f'{ERROR} Problem while joining price with cents - {self.name}')
# Return data
if self.DEBUG == True:
print([self.p_name, self.price, self.currency])
data_collected = ShopScrapersItems()
data_collected['url'] = response.url
data_collected['p_name'] = self.p_name
data_collected['price'] = self.price
data_collected['currency'] = self.currency
yield data_collected
通常,当我从控制台运行蜘蛛时,我会:
scrapy crawl ElCorteIngles -a url='https://www.elcorteingles.pt/electrodomesticos/A26601428-depiladora-braun-senso-smart-5-5500/'
现在我需要一种在外部脚本上执行相同操作并获取输出的方法yield data_collected
我目前在我的外部脚本中拥有的是:
import scrapy
from scrapy.crawler import CrawlerProcess
import sqlalchemy as db
# Import internal libraries
from Ruby.Ruby.spiders import *
# Variables
engine = db.create_engine('mysql+pymysql://DATABASE_INFO')
class Worker(object):
def __init__(self):
self.crawler = CrawlerProcess({})
def scrape_new_links(self):
conn = engine.connect()
# Get all new links from DB and scrape them
query = 'SELECT * FROM Ruby.New_links'
result = conn.execute(query)
for x in result:
telegram_id = x[1]
email = x[2]
phone_number = x[3]
url = x[4]
spider = x[5]
# In this cade the spider will be ElCorteIngles and
# the url https://www.elcorteingles.pt/electrodomesticos/A26601428-depiladora-
# braun-senso-smart-5-5500/'
self.crawler.crawl(spider, url=url)
self.crawler.start()
Worker().scrape_new_links()
我也不知道这样做是否url=url
是self.crawler.crawl()
将 URL 提供给蜘蛛的正确方法,但请告诉我您的想法。来自的所有数据yield
都由管道返回。我认为不需要额外的信息,但如果您需要任何信息,请告诉我!
解决方案
Scrapy 异步工作...忽略我的导入,但这是我为 scrapy 制作的 JSON api。您需要使用 item_scraped 信号制作自定义跑步者。最初有一个 klein 端点,当蜘蛛完成时,它会返回一个 JSON 列表。我认为这就是你想要的,但没有克莱因端点,所以我把它拿出来了。我的蜘蛛是 GshopSpider 我用你的蜘蛛名字代替了它。
通过利用 deferred,我们可以在每次抓取项目时使用回调并发送信号。所以使用这段代码,我们将每个项目收集到一个带有信号的列表中,当蜘蛛完成时,我们有一个回调设置到 return_spider_output
# server.py
import json
from scrapy import signals
from scrapy.crawler import CrawlerRunner
from googleshop.spiders.gshop import GshopSpider
from scrapy.utils.project import get_project_settings
class MyCrawlerRunner(CrawlerRunner):
def crawl(self, crawler_or_spidercls, *args, **kwargs):
# keep all items scraped
self.items = []
crawler = self.create_crawler(crawler_or_spidercls)
crawler.signals.connect(self.item_scraped, signals.item_scraped)
dfd = self._crawl(crawler, *args, **kwargs)
dfd.addCallback(self.return_items)
return dfd
def item_scraped(self, item, response, spider):
self.items.append(item)
def return_items(self, result):
return self.items
def return_spider_output(output):
return json.dumps([dict(item) for item in output])
if __name__=="__main__"
settings = get_project_settings()
runner = MyCrawlerRunner(settings)
spider = ElCorteIngles()
deferred = runner.crawl(spider)
deferred.addCallback(return_spider_output)
return deferred
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