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

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"
    }
}
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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 …
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python error-handling scrapy scrapy-splash

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