Swift 2 中的 HTTP POST 错误处理

标签 http post error-handling nsjsonserialization swift2

我是新来的,这是我的第一个问题... 我尝试在 Swift 2 中编写一个发出 HTTP POST 请求的应用程序,但我不知道如何使用 Swift 2 的新错误处理。谁能告诉我如何实现“do-try- catch"Swift 2 的错误处理到下面的代码片段? (此代码片段使用 swift 1.2 的旧错误处理)

func post(params : Dictionary<String, String>, url : String) {
    var request = NSMutableURLRequest(URL: NSURL(string: url)!)
    var session = NSURLSession.sharedSession()
    request.HTTPMethod = "POST"

    var err: NSError?
    request.HTTPBody = NSJSONSerialization.dataWithJSONObject(params, options: nil/*, error: &err*/)
    request.addValue("application/json", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept")

    var task = session.dataTaskWithRequest(request, completionHandler: {data, response, error -> Void in
        print("Response: \(response)")
        var strData = NSString(data: data!, encoding: NSUTF8StringEncoding)
        print("Body: \(strData)")
        var err: NSError?
        var json = NSJSONSerialization.JSONObjectWithData(data!, options: .MutableLeaves/*, error: &err*/) as? NSDictionary

        // Did the JSONObjectWithData constructor return an error? If so, log the error to the console
        if(err != nil) {
            print(err!.localizedDescription)
            let jsonStr = NSString(data: data!, encoding: NSUTF8StringEncoding)
            print("Error could not parse JSON: '\(jsonStr)'")
        }
        else {
            // The JSONObjectWithData constructor didn't return an error. But, we should still
            // check and make sure that json has a value using optional binding.
            if let parseJSON = json {
                // Okay, the parsedJSON is here, let's get the value for 'success' out of it
                var success = parseJSON["success"] as? Int
                print("Succes: \(success)")
            }
            else {
                // Woa, okay the json object was nil, something went worng. Maybe the server isn't running?
                let jsonStr = NSString(data: data!, encoding: NSUTF8StringEncoding)
                print("Error could not parse JSON: \(jsonStr)")
            }
        }
    })

    task!.resume()
}

最佳答案

您可能希望将您的 NSJSONSerialization 调用包装在 do/try/catch 逻辑中,如下所示。

在 Swift 3 中:

var request = URLRequest(url: URL(string: urlString)!)

let session = URLSession.shared
request.httpMethod = "POST"
request.addValue("application/json", forHTTPHeaderField: "Content-Type")
request.addValue("application/json", forHTTPHeaderField: "Accept")

request.httpBody = try! JSONSerialization.data(withJSONObject: parameters)

// or if you think the conversion might actually fail (which is unlikely if you built `parameters` yourself)
//
// do {
//    request.httpBody = try JSONSerialization.data(withJSONObject: parameters)
// } catch {
//    print(error)
// }

let task = session.dataTask(with: request) { data, response, error in
    guard let data = data, error == nil else {
        print("error: \(error)")
        return
    }

    // this, on the other hand, can quite easily fail if there's a server error, so you definitely
    // want to wrap this in `do`-`try`-`catch`:

    do {
        if let json = try JSONSerialization.jsonObject(with: data) as? [String: Any] {
            let success = json["success"] as? Int                                  // Okay, the `json` is here, let's get the value for 'success' out of it
            print("Success: \(success)")
        } else {
            let jsonStr = String(data: data, encoding: .utf8)    // No error thrown, but not dictionary
            print("Error could not parse JSON: \(jsonStr)")
        }
    } catch let parseError {
        print(parseError)                                                          // Log the error thrown by `JSONObjectWithData`
        let jsonStr = String(data: data, encoding: .utf8)
        print("Error could not parse JSON: '\(jsonStr)'")
    }
}

task.resume()

或者,在 Swift 2 中

let request = NSMutableURLRequest(URL: NSURL(string: urlString)!)

let session = NSURLSession.sharedSession()
request.HTTPMethod = "POST"
request.addValue("application/json", forHTTPHeaderField: "Content-Type")
request.addValue("application/json", forHTTPHeaderField: "Accept")

request.HTTPBody = try! NSJSONSerialization.dataWithJSONObject(parameters, options: [])

// or if you think the conversion might actually fail (which is unlikely if you built `parameters` yourself)
//
// do {
//    request.HTTPBody = try NSJSONSerialization.dataWithJSONObject(params, options: [])
// } catch {
//    print(error)
// }

let task = session.dataTaskWithRequest(request) { data, response, error in
    guard let data = data where error == nil else {
        print("error: \(error)")
        return
    }

    // this, on the other hand, can quite easily fail if there's a server error, so you definitely
    // want to wrap this in `do`-`try`-`catch`:

    do {
        if let json = try NSJSONSerialization.JSONObjectWithData(data, options: []) as? NSDictionary {
            let success = json["success"] as? Int                                  // Okay, the `json` is here, let's get the value for 'success' out of it
            print("Success: \(success)")
        } else {
            let jsonStr = String(data: data, encoding: NSUTF8StringEncoding)    // No error thrown, but not NSDictionary
            print("Error could not parse JSON: \(jsonStr)")
        }
    } catch let parseError {
        print(parseError)                                                          // Log the error thrown by `JSONObjectWithData`
        let jsonStr = String(data: data, encoding: NSUTF8StringEncoding)
        print("Error could not parse JSON: '\(jsonStr)'")
    }
}

task.resume()

我还建议对 data 的强制展开更加小心,因为您想要检测/处理错误,而不是崩溃。例如,上面我使用了 guard 语句来解包它。

关于Swift 2 中的 HTTP POST 错误处理,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31030366/

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