如何使用scrapysplash从该网站获取dinamicallyloaded内容？

Question

我正在尝试使用scrapy splash从这个website获取数据，但我无法提取数据。我想获取每个真实状态的数据，如href、price等。以下是我的代码：

在settings.py中：

ROBOTSTXT_OBEY = False

USER_AGENT = "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/34.0.1847.131 Safari/537.36"

SPLASH_ENABLED = True


DOWNLOADER_MIDDLEWARES = {
    'scrapy_splash.SplashCookiesMiddleware': 723,
    'scrapy_splash.SplashMiddleware': 725,
    'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
}

SPLASH_URL = 'http://localhost:8050/'

SPIDER_MIDDLEWARES = {
    'scrapy_splash.SplashDeduplicateArgsMiddleware': 100,
}

DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter'
HTTPCACHE_STORAGE = 'scrapy_splash.SplashAwareFSCacheStorage'

我的蜘蛛：

class M2Spider(scrapy.Spider):
    
    name = "m2"
    allowed_domains = ['metrocuadrado.com']
    start_urls = [
            'https://www.metrocuadrado.com/bodega/arriendo'
            ]
    
    def start_requests(self):
        for url in self.start_urls:
            yield SplashRequest(url=url,callback= self.parse, 
                               endpoint='render.html',
                               args={'wait': 10},)
                  
    def parse(self, response):
        print("--------------------------------------------------------------")
        real_states= response.selector.xpath(".//a[@class='sc-bdVaJa ebNrSm']").getall()

        print("real_states")

输出打印为空列表[]。我是个新手。有什么建议吗

Answer 1

我要做的是：

向https://www.metrocuadrado.com/results/_next/static/chunks/commons.8afec6af6d5add2097bf.js发送请求，如果搜索“X-API-key”，您将在响应中找到一个API键。这样就可以很容易地用正则表达式提取，比如：re.findall(r'"X-Api-Key":"(\w+)"')

然后，提取API密钥后，向https://www.metrocuadrado.com/rest-search/search?seo=/bodega/arriendo&from=0&size=50发送请求，这是您发送的网站中隐藏的API。要获得有效的响应，必须像这样附加标题

scrapy.Request(
    url=url_variable,
    headers={
        "x-api-key": api_key_variable_from_prev_step
    }
)

通过该API，您可以获得JSON格式的数据，这通常比解析html更可靠，因为这会更频繁地发生变化