【问题标题】:Swift - Encoding keys not being properly sent in HTTP POSTSwift - 在 HTTP POST 中未正确发送编码密钥
【发布时间】:2019-08-02 18:24:01
【问题描述】:

我正在我的应用程序中处理 POST 请求,该请求会将有关照片的各种信息发送到服务器(注意:我不拥有服务器)。为了获得成功响应,我需要发送一个 JSON 对象,其中包含该照片的以下键及其相关值。

{
    "description":"example",
    "media_url":"https://example.com",
    "attribute[request_type]":"Example",
    "service_code":"123456789",
    "lat":"0.0",
    "api_key":"987654321",
    "long":"0.0"
}

在此调用中,每个键名都需要完全匹配。

我的问题在于属性[request_type] 键。由于某种原因,当我通过我的应用程序发送它时,它没有正确识别该键名。我知道这一点,因为它给了我以下响应消息

    [{"code":400,"description":"Attribute request_type required"}]

我知道它所说的键名与 JSON 对象中的键名不同,但我使用 Postman 测试过,属性 [request_type] 是正确的命名约定。

我知道问题不在于我传入的值,否则它会告诉我属性 request_type 无效。

当我打印 JSON 对象时,所有字段都在那里。

控制台的确切输出:

    {"description":"example","media_url":"https:\/\/example.com\/","attribute[request_type]":"example","service_code":"987654321","lat":"0.0","api_key":"123456789","long":"0.0"}

这就是我设置结构的方式

struct RequestModel: Codable {
    var api_key: String
    var service_code: String
    var description: String
    var media_url: String
    var requestType: String
    var long: String
    var lat: String

    enum CodingKeys: String, CodingKey {
        case requestType = "attribute[request_type]"

        case api_key
        case service_code
        case description
        case long
        case lat
        case media_url
    }

    init(api_key: String, service_code: String, lat: String, long: String, media_url: String, description: String, requestType: String) {
        self.api_key = api_key
        self.service_code = service_code
        self.description = description
        self.media_url = media_url
        self.long = long
        self.lat = lat
        self.requestType = requestType
    }
}

这就是函数本身

func send(lat: Double, long: Double, comment: String, photoURL: String) {
        let lattitude = String(lat)
        let longitute = String(long)
        let description = comment
        let mediaURL = photoURL

        //Prepping Data
        let sendReqeust = RequestModel(api_key: "123456789", service_code: "987654321", lat: lattitude, long: longitute, media_url: mediaURL, description: description, requestType: "example")

        guard let uploadData = try? JSONEncoder().encode(sendReqeust) else {
            return
        }
        print(String(data: uploadData, encoding: .utf8)!)

        //Configuring an Upload Request
        let url = URL(string: "http://example.com")!
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        request.setValue("application/json", forHTTPHeaderField: "Content-Type")

        //Creating and Starting an Upload Task
        let task = URLSession.shared.uploadTask(with: request, from: uploadData) { data, response, error in
            print("Response: \(response)")
            if let error = error {
                print ("error: \(error)")
                return
            }
            guard let response = response as? HTTPURLResponse,
                (200...299).contains(response.statusCode) else {
                    print(String(data: data!, encoding: .utf8)!)
                    print ("server error")
                    return
            }
            if let mimeType = response.mimeType,
                mimeType == "application/json",
                let data = data,
                let dataString = String(data: data, encoding: .utf8) {
                print ("got data: \(dataString)")
            }
        }
        task.resume()
    }

我没有正确设置编码密钥吗?我的编码方式有问题吗?

【问题讨论】:

  • 嘿!不清楚,它是否适用于Postman?如果是,请附上cURLsn-p?我们需要确保它不是服务器端错误。为此,我们可以使用任何REST 客户端工具。如何从Postman 获取cURL snipper 你可以在这里找到learning.getpostman.com/docs/postman/sending_api_requests/…
  • @AlexD。我可以报告它在邮递员中确实有效,但是我不能在不暴露 api_key 的情况下提供 cURL。

标签: json swift post


【解决方案1】:

我在处理Alamofire 令牌时遇到了类似的问题,我用String concat 解决了。

试试这个:

 request.addValue("{\"request_type\":\"Example\"}", forHTTPHeaderField: "attribute")

编辑

如果确实需要将信息传递到请求正文中,请尝试如下解析:

{
    "description":"example",
    "media_url":"https://example.com",
    "attribute": {
        "request_type":"Example"
    },
    "service_code":"123456789",
    "lat":"0.0",
    "api_key":"987654321",
    "long":"0.0"
}

解决方案

我遵循了上面的建议,所以现在代码看起来像这样

    // in the struct requestType now looks like this
    var attribute: [String: String]
    // in the init within the function looks like this
    let sendReqeust = RequestModel(api_key: "987654321", service_code: "123456789", lat: lattitude, long: longitute, media_url: mediaURL, description: description, attribute: ["request_type":"example"])

【讨论】:

  • 谢谢!这最终完美运行,提交了一个包含必要代码更改的编辑以供大家将来参考
【解决方案2】:

将 requestType = "attribute[request_type]" 更改为 requestType = "request_type"

【讨论】:

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