Scrapy-Splash ERROR 400：“描述”：“缺少必需的参数：url”

Question

我在我的代码中使用scrapy splash 来生成javascript-html 代码。

飞溅把这个 render.html 还给我

{
    "error": 400,
    "type": "BadOption",
    "description": "Incorrect HTTP API arguments",
    "info": {
        "type": "argument_required",
        "argument": "url",
        "description": "Required argument is missing: url"
    }
}

而且我无法通过 javascript 生成的 html 获得响应。这是我的spider.py

class ThespiderSpider(scrapy.Spider):
    name = 'thespider'
    #allowed_domains = ['https://www.empresia.es/empresa/repsol/']
    start_urls = ['https://www.empresia.es/empresa/repsol/']

    def start_requests(self):
        yield scrapy.Request( 'http://example.com', self.fake_start_requests )

    def fake_start_requests(self, response):
        for url in self.start_urls:
            yield SplashRequest( url, self.parse,
                                 args={'wait': 1.5, 'http_method': 'POST'},
                                 endpoint='render.html'
                                 )

    def parse(self, response):
        open_in_browser(response)
        title = response.css("title").extract()
        # har = response.data["har"]["log"]["pages"]
        headers = response.headers.get('Content-Type')
        names = response.css('.fa-user-circle-o+ a::text').extract()
        yield {
            'title': title,
            #'har': har,
            'headers': headers,
            'names': names,
            'length': len(names)
        }

这是我的 settings.py

USER_AGENT = 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/79.0.3945.88 Safari/537.36'

# Obey robots.txt rules
ROBOTSTXT_OBEY = True

# Splash Settings
DOWNLOADER_MIDDLEWARES = {
    # Engine side
    'scrapy_splash.SplashCookiesMiddleware': 723,
    'scrapy_splash.SplashMiddleware': 725,
    'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
    # Downloader side
}

SPIDER_MIDDLEWARES = {
    'scrapy_splash.SplashDeduplicateArgsMiddleware': 100,
}
SPLASH_URL = 'http://127.0.0.1:8050/'
# SPLASH_URL = 'http://192.168.59.103:8050/'
DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter'
HTTPCACHE_STORAGE = 'scrapy_splash.SplashAwareFSCacheStorage'

谢谢您的帮助。

Answer 1

url在参数中提供参数args即可工作。

yield SplashRequest(self.parse,
                     args={'url': url, 'wait': 1.5, 'http_method': 'POST'},
                     endpoint='render.html'
                     )