Ios 状态代码应为200，但为415

我试图发布一个简单的字符串，但我不断收到HTTP 415不支持的媒体类型错误。我尝试将参数转换为JSON，但仍然没有成功

方法更新

func requestUsingPostMethod(url: String, parameter: String, completion: @escaping (_ success: [String : AnyObject]) -> Void) {

    //@escaping...If a closure is passed as an argument to a function and it is invoked after the function returns, the closure is @escaping.

    var request = URLRequest(url: URL(string: url)!)
    request.addValue("application/json", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept")
    request.httpMethod = "POST"
    let postString = parameter


    request.httpBody = try? JSONSerialization.data(withJSONObject: [postString])
      //  request.httpBody = postString.data(using: .utf8)
    let task = URLSession.shared.dataTask(with: request) { Data, response, error in

        guard let data = Data, error == nil else {  // check for fundamental networking error

            print("error=\(String(describing: error))")
            return
        }

        if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 {  // check for http errors

            print("statusCode should be 200, but is \(httpStatus.statusCode)")
            print(response!)
            return

        }

        let responseString  = try! JSONSerialization.jsonObject(with: data, options: .allowFragments) as! [String : AnyObject]
        completion(responseString)


    }

    task.resume()

}

请求

NetworkCall().requestUsingPostMethod(url: "http://192.168.50.119:8181/rest/items/wemo_lamp_switch", parameter: "ON", completion: { response in

            print("--------------------------------------------------------------")
            print(response)
           // let jsonResults = JSON(String: response)
        })

错误

statusCode应为200，但为415{URL: }{地位 代码：415，页眉{ “内容长度”=282； “内容类型”=“应用程序/json”； 日期=“2017年9月12日星期二08:33:16 GMT”； Server=“Jetty（9.2.19.v20160908）”；}

我用你的答案更新了问题，但仍然得到相同的错误

邮差数据

{
    "id": "6f5d4f8a-612a-10f9-71b5-6dc8ba668885",
    "name": "simpledata",
    "description": "",
    "order": [
        "65df8736-1069-b0f0-3a1d-c318ce1810e0"
    ],
    "folders": [],
    "folders_order": [],
    "timestamp": 1505208950131,
    "owner": 0,
    "public": false,
    "requests": [
        {
            "id": "65df8736-1069-b0f0-3a1d-c318ce1810e0",
            "headers": "",
            "headerData": [],
            "url": "http://192.168.50.119:8181/rest/items/wemo_lamp_switch",
            "queryParams": [],
            "pathVariables": {},
            "pathVariableData": [],
            "preRequestScript": null,
            "method": "POST",
            "collectionId": "6f5d4f8a-612a-10f9-71b5-6dc8ba668885",
            "data": [],
            "dataMode": "raw",
            "name": "http://192.168.50.119:8181/rest/items/wemo_lamp_switch",
            "description": "",
            "descriptionFormat": "html",
            "time": 1505208951279,
            "version": 2,
            "responses": [],
            "tests": null,
            "currentHelper": "normal",
            "helperAttributes": {},
            "rawModeData": "OFF"
        }
    ]
}

---

将请求的

内容类型

标题设置为您希望API返回的任何内容。如果您确定它返回一个

JSON

，那么

request.allHTTPHeaderFields["Content-Type"] = "application/json"

也

---

您正在向API发送UTF8编码的字符串，而不是JSON，因此出现HTTP错误415。但是，一个简单的字符串不能转换为JSON，它需要是数组或字典的一部分，因此您需要确定API所期望的实际格式

request.httpBody=try？JSONSerialization.data（带jsonObject:[postString]）

您可能还需要将

内容类型

头添加到HTTP头中

---

请求。allHTTPHeaderFields[“内容类型”]=“应用程序/json”

415表示“不支持的媒体类型”。您可能需要将accept内容类型添加到

request

中的“application/json”？您正在向WS-waiting发送一个字符串，等待JSONhy tnx。即使加上那个。我得到了同样的错误。即使加上那个。我收到了相同的错误。@“application/json；charset=UTF-8”尝试添加编码alsohy tnx。即使加上那个。我也遇到了同样的错误。你真的设法让你的API在其他平台上工作了吗？您应该首先让请求在测试平台（如Postman）上工作，并将工作请求包含在问题中。看起来，您实际上不知道API需要什么格式。@David.tnx。我试着用邮递员作为原始数据。它很好用。用邮递员数据更新了我的问题，请检查。@SathyaBaman这是请求的正文还是响应的标题？您应该在问题中包括您在Postman中使用的任何标题、包含任何查询参数（如果使用）的完整URL以及请求正文。@“application/json；charset=UTF-8”请尝试添加编码

request.addValue("application/json", forHTTPHeaderField: "Content-Type")
request.addValue("application/json", forHTTPHeaderField: "Accept")

let theparam = JSONSerialization.data(withJSONObject: parameter)