如何使用 scrapy-splash 从该网站获取动态加载的内容?
How to get dinamically-loaded content from this website using scrapy-splash?
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。我想获取有关每个真实状态的数据,如 href、价格等。这是我的代码:
在setings.py:
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")
输出打印是一个空列表[]。我是新手。有什么建议吗?
我会做的是:
向 https://www.metrocuadrado.com/results/_next/static/chunks/commons.8afec6af6d5add2097bf.js 发送请求,如果您搜索“X-Api-Key”,您会在响应中找到 API-key。这样就可以使用正则表达式轻松提取,例如:re.findall(r'"X-Api-Key":"(\w+)"')
.
然后,当您提取了 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。要获得有效回复,您必须像这样附加 header
scrapy.Request(
url=url_variable,
headers={
"x-api-key": api_key_variable_from_prev_step
}
)
从 API 你得到 JSON 格式的数据,这通常比解析 html 更可靠,因为它经常改变。
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。我想获取有关每个真实状态的数据,如 href、价格等。这是我的代码:
在setings.py:
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")
输出打印是一个空列表[]。我是新手。有什么建议吗?
我会做的是:
向 https://www.metrocuadrado.com/results/_next/static/chunks/commons.8afec6af6d5add2097bf.js 发送请求,如果您搜索“X-Api-Key”,您会在响应中找到 API-key。这样就可以使用正则表达式轻松提取,例如:re.findall(r'"X-Api-Key":"(\w+)"')
.
然后,当您提取了 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。要获得有效回复,您必须像这样附加 header
scrapy.Request(
url=url_variable,
headers={
"x-api-key": api_key_variable_from_prev_step
}
)
从 API 你得到 JSON 格式的数据,这通常比解析 html 更可靠,因为它经常改变。