首先,我获得了一个zomato api . 然后我使用nodejs fetch从api获取餐馆 .
0-19项:
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=0&count=20
20-40项:
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=20&count=20
40-60项:
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=40&count=20
60-80项:
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=60&count=20
80-100项:
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=80&count=20
100-120(从此开始,它开始返回空记录)
https://developers.zomato.com/api/v2.1/search?entity_id=259&entity_type=city&start=100&count=20
我的问题是,它是一种返回更多项目的方法 .
1 回答
对于特定搜索条件,上限限制为100个结果 . 无法通过搜索API获取超过100个商家信息(这适用于基本和合作伙伴API访问) .