web-services - 如何使用 NSURLSession 在 Swift 中定义内容类型

Question

我想在下面的代码中设置内容类型来调用 web api。内容类型将为 application/json； charset=utf-8

let url = NSURL(string: "http:/api/jobmanagement/PlusContactAuthentication?email=\(usr)&userPwd=\(pwdCode)")

println("URL:  \(url)")

let task = NSURLSession.sharedSession().dataTaskWithURL(url!) {
    (data, response, error) in
    println(NSString(data: data, encoding: NSUTF8StringEncoding))
}

//  task.setValue(<#value: AnyObject?#>, forKey: <#String#>)
task.resume()

Answer 1

如果你想设置请求的Content-Type，你可以创建你自己的URLRequest，提供你的URL，指定Content-Type 使用 setValue(_:forHTTPHeaderField:) header ，然后使用 URLRequest 而不是直接使用 URL 发出请求。只需将 httpBody 设置为该 JSON 并指定 POST 的 httpMethod:

let url = URL(string: "https://api/jobmanagement/PlusContactAuthentication")!
var request = URLRequest(url: url)
request.setValue("application/json; charset=utf-8", forHTTPHeaderField: "Content-Type")  // the request is JSON
request.setValue("application/json; charset=utf-8", forHTTPHeaderField: "Accept")        // the expected response is also JSON
request.httpMethod = "POST"

let dictionary = ["email": username, "userPwd": password]
request.httpBody = try! JSONEncoder().encode(dictionary)

let task = URLSession.shared.dataTask(with: request) { data, response, error in
    guard let data = data, error == nil else {
        print(error ?? "Unknown error")                                 // handle network error
        return
    }

    // parse response; for example, if JSON, define `Decodable` struct `ResponseObject` and then do:
    //
    // do {
    //     let responseObject = try JSONDecoder().decode(ResponseObject.self, from: data)
    //     print(responseObject)
    // } catch let parseError {
    //     print(parseError)
    //     print(String(data: data, encoding: .utf8))   // often the `data` contains informative description of the nature of the error, so let's look at that, too
    // }
}
task.resume()

有关 Swift 2 版本，请参阅 previous revision of this answer .