ios – 如何在Swift中优化链接异步请求

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我的代码执行请求链接像A-> B-> C我正在使用URLSession所有请求都按正确的顺序完成并具有预期的行为.但我想知道如何优化这种链接,因为它看起来非常复杂而且不可重复使用.我正在寻找如何以更灵活的方式进行链接的建议.

我的代码

URLSession.shared.dataTask(with: URLRequest(url: URL(string: "first")!)){ data,res,err in
    let second = URLRequest(url: URL(string: "second")!)
    URLSession.shared.dataTask(with: second){ data,err in
        let third = URLRequest(url: URL(string: "second")!)
        URLSession.shared.dataTask(with:third){ data,err in
         }.resume()
    }.resume()
}.resume()

解决方法

正如@Paulw11所说:

PromiseKit PMKFoundation

import PromiseKit
import PMKFoundation

let session = URLSession.shared
firstly {
    session.dataTask(with: URLRequest(url: URL(string: "first")!))
} .then { data in
    session.dataTask(with: URLRequest(url: URL(string: "second")!))
} .then { data in
    session.dataTask(with: URLRequest(url: URL(string: "third")!))
} .then { data -> () in
    // The data here is the data fro the third URL
} .catch { error in
    // Any error in any step can be handled here
}

使用1(且仅1)重试,您可以使用恢复.恢复就像捕获,但预计之前的那个可以重试.但是,这不是一个循环,只执行一次.

func retry(url: URL,on error: Error) -> Promise<Data> {
    guard error == MyError.retryError else { throw error }

    // Retry the task if a retry-able error occurred.
    return session.dataTask(with: URLRequest(url: url))
}

let url1 = URL(string: "first")!
let url2 = URL(string: "second")!
let url3 = URL(string: "third")!

let session = URLSession.shared
firstly {
    session.dataTask(with: URLRequest(url: url1))
} .then { data in
    session.dataTask(with: URLRequest(url: url2))
} .recover { error in
    retry(url: url2,on: error)
} .then { data in
    session.dataTask(with: URLRequest(url: url3))
} .recover { error in
    retry(url: url3,on: error)
} .then { data -> () in
    // The data here is the data fro the third URL
} .catch { error in
    // Any error in any step can be handled here
}

注意:为了使这项工作不指定返回类型并需要一个return语句,我需要then然后恢复为1行.所以我创建了进行处理的方法.

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