Ios 在swift 3中,如何将从服务器获得的响应从字符串转换为数组?

Ios 在swift 3中,如何将从服务器获得的响应从字符串转换为数组?,ios,json,swift,Ios,Json,Swift,在这里,我在成功发布参数后得到了响应,我需要将其检索回来,但我遇到的问题是,我已将数据保存在responseString中,并且它以字符串的形式存储,当我尝试检索数据并保存在数组中时,无法保存,有人能帮助我如何保存数据吗,数据在下面格式 这是服务器响应 [ { "carrier_code": "flatrate", "method_code": "flatrate", "carrier_title": "Flat Rate", "method_title":

在这里,我在成功发布参数后得到了响应,我需要将其检索回来,但我遇到的问题是,我已将数据保存在
responseString
中,并且它以字符串的形式存储,当我尝试检索数据并保存在数组中时,无法保存,有人能帮助我如何保存数据吗,数据在下面格式

这是服务器响应

[
  {
    "carrier_code": "flatrate",
    "method_code": "flatrate",
    "carrier_title": "Flat Rate",
    "method_title": "Fixed",
    "amount": 0,
    "base_amount": 0,
    "available": true,
    "error_message": "",
    "price_excl_tax": 0,
    "price_incl_tax": 0
  },
  {
    "carrier_code": "tablerate",
    "method_code": "bestway",
    "carrier_title": "Best Way",
    "method_title": "Table Rate",
    "amount": 0,
    "base_amount": 0,
    "available": true,
    "error_message": "",
    "price_excl_tax": 0,
    "price_incl_tax": 0
  }
]
下面是用于发布参数的json函数

        func shippingmethodURL(shippingMethodAPI:String) {
        let url = NSURL(string: shippingMethodAPI)
        var request = URLRequest(url: url! as URL)
        request.httpMethod = "POST"
        print(shippingMethodAPI)
        let storyboard = UIStoryboard(name: "Main", bundle: nil)
        let addtoCartVC = storyboard.instantiateViewController(withIdentifier: "checkout") as! CheckoutViewController
        let parameters : [String: Any] = ["address":
            [ "region": "California",
                "region_code": "CA",
                "region_id": "12",
                "country_id": "US",
                "company": "Test",
                "telephone": "9492162752",
                "postcode": "43",
                "city": "Chennai",
                "firstname": "gdfgdgdfg",
                "lastname": "dgdfgdfgg",
                "email": "sfdsfsdf@gmail.com",
                "prefix": "",
                "sameAsBilling": 1,
                "street": ["Dsfdsfsd dfdsfdsf dsfsfdsfsf sdfsfdsfsdfC"]]]
        print(parameters)
        do {
            request.httpBody = try JSONSerialization.data(withJSONObject: parameters, options: .prettyPrinted)

        } catch let error {
            print(error.localizedDescription)
        }
        request.addValue("application/json", forHTTPHeaderField: "Content-Type")
        print(request)
        let task = URLSession.shared.dataTask(with: request) { data, response, error in
            guard let data = data, error == nil else {
                print("error=\(String(describing: error))")
                return
            }

            if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 {
                print("statusCode should be 200, but is \(httpStatus.statusCode)")
                print("response = \(String(describing: response))")
            }
            let responseString = String(data: data, encoding: .utf8)
            print("responseString = \(responseString!)")

            let status = (response as! HTTPURLResponse).statusCode
            self.keyStatusCode = status
            print(status)
            let array = responseString
        }
        task.resume()
    }

您正在将
数据
转换为
字符串
,但是它是
数组
您需要使用
JSONSerialization
类来实现这一点

您必须替换此代码

 let responseString = String(data: data, encoding: .utf8)

编辑

您需要将其放入
中,并尝试像这样捕获

   do {
        let array = try JSONSerialization.jsonObject(with: data) as? [[String : Any]]

    } catch {
        print("Exception occured \(error))")
    }

您需要有一个模型类,然后有一个该模型类的数组。现在您可以将获取的对象保存在数组中。如何将输出服务器数据连接到模型类@Tushars与此相关,但服务器没有美感。它根本不关心漂亮的打印文本。你需要把它放在do try catch BlockThank working中如果我需要检索数据,那么我需要为它创建模型类@Jon Snower如果你使用Alamofire对象映射器(第三方库),它会自动为你做,您可以使用
SwiftyJSONAccelerator
(Mac app)@LeoDabus编辑的应答器创建对象映射器类注意,添加本地化描述是多余的<代码>打印(“json错误:,错误)
   do {
        let array = try JSONSerialization.jsonObject(with: data) as? [[String : Any]]

    } catch {
        print("Exception occured \(error))")
    }