从json中的脚本输出中抓取

Question

我scrapy在python脚本中运行

def setup_crawler(domain):
    dispatcher.connect(stop_reactor, signal=signals.spider_closed)
    spider = ArgosSpider(domain=domain)
    settings = get_project_settings()
    crawler = Crawler(settings)
    crawler.configure()
    crawler.crawl(spider)
    crawler.start()
    reactor.run()

它成功运行并停止，但结果在哪里？我想要json格式的结果，该怎么办？

result = responseInJSON

就像我们使用命令一样

scrapy crawl argos -o result.json -t json

Answer 1

您需要手动设置FEED_FORMAT和FEED_URI设置：

settings.overrides['FEED_FORMAT'] = 'json'
settings.overrides['FEED_URI'] = 'result.json'

如果要将结果放入变量中，可以定义一个Pipeline将项目收集到列表中的类。使用spider_closed信号处理程序查看结果：

import json

from twisted.internet import reactor
from scrapy.crawler import Crawler
from scrapy import log, signals
from scrapy.utils.project import get_project_settings


class MyPipeline(object):
    def process_item(self, item, spider):
        results.append(dict(item))

results = []
def spider_closed(spider):
    print results

# set up spider    
spider = TestSpider(domain='mydomain.org')

# set up settings
settings = get_project_settings()
settings.overrides['ITEM_PIPELINES'] = {'__main__.MyPipeline': 1}

# set up crawler
crawler = Crawler(settings)
crawler.signals.connect(spider_closed, signal=signals.spider_closed)
crawler.configure()
crawler.crawl(spider)

# start crawling
crawler.start()
log.start()
reactor.run()

仅供参考，看看Scrapy如何解析命令行参数。