ajax回调地狱解决方案

Posted 2022-04-20 Fire king

//1.封装
function getData(url , data = ) 
	return new Promise((resolve , reject) = > 
		$.ajax(
			//发送请求

			data: data
			sussess: function (res) 
				resolve(res)
			,
			error: function (res) 
				reject(res)
			
		)
	)


//2.使用
getData(url1)
	.then((res) => 
		const id = res
		return getData(url2 , id)
	)
		.then((res) => 
		const name = res
		return getData(url3 , name)
	)
			.then((res) => 
		console.log(res)
	)
.catch((err) => 
	console.log(err)
)