Question

我正在使用NSURL从网站上抓取HTML。问题是，如果在NSURL请求期间互联网中断，应用程序崩溃。
代码：

let myUrl = NSURL(string: "http://www.mywebsite.com/page.html")
let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
    data, response, error in
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0)) {
        let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
        if error != nil {
             print("Error: \(error)")
        } 
        dispatch_async(dispatch_get_main_queue()) {
            self.testLabel.text = "\(responseString!)"
        }
    }
}
task.resume()

我可以在使用此代码进行NSURL连接之前检查互联网连接，但是在操作过程中仍有可能断开互联网：
在NSURL之前检查网络连接的代码：

let myUrl = NSURL(string: "http://www.mywebsite.com/page.html")
let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
    data, response, error in
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0)) {
        let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
        if error != nil {
             print("Error: \(error)")
        } 
        dispatch_async(dispatch_get_main_queue()) {
            self.testLabel.text = "\(responseString!)"
        }
    }
}
task.resume()

为了解决这个问题我已经研究过使用try / catch操作，但是我没有成功构建一个不会给我错误的运气。有没有办法将整个NSURL操作包装在一个捕获任何错误的巨型try / catch中？我正在思考这方面的问题，但它不起作用：
代码：

let myUrl = NSURL(string: "http://www.mywebsite.com/page.html")
let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0)) {
    let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
    if error != nil {
         print("Error: \(error)")
    } 
    dispatch_async(dispatch_get_main_queue()) {
        self.testLabel.text = "\(responseString!)"
    }
  }
}
task.resume()
} catch {
    timer.invalidate
    sendAlert("Error", message: "You are not connected to the internet")
}

Answer 1

!发音的正确方法是＆＃34;我发誓我的节目生活。＆＃34;因为这就是你正在做的事情。为什么你觉得必须使用!？使用guard-let为您解决所有问题：

    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in

        guard error == nil else {
            print("Error: \(error)")
            return
        }

        guard let data = data else {
            fatalError("error was nil, but no data. This should never happen")
        }

        guard let responseString = NSString(data: data, encoding: NSUTF8StringEncoding) else {
            print("Couldn't decode data")
            return
        }

        dispatch_async(dispatch_get_main_queue()) {
            self.testLabel.text = "\(responseString)"
        }
    }

Answer 2

如果你知道100％有一个值，你应该只强制解包变量，否则你会导致你的应用崩溃。如果您需要更多相关信息，请查看The Swift Programming Language guide

的可选链接部分

检查

以下代码适合您。

    let myUrl = NSURL(string: "http://www.mywebsite.com/page.html")
    let request = NSMutableURLRequest(URL: myUrl!)
    request.HTTPMethod = "POST"
    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in
        dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0)) {

            if let responseData = data {

                let responseString = NSString(data: responseData, encoding: NSUTF8StringEncoding)
                if error != nil {
                    print("Error: \(error)")
                }
                dispatch_async(dispatch_get_main_queue()) {
                    print("Response: \(responseString!)")
                }
            }

        }
    }

    task.resume()

在处理您的互联网连接失败方面，您应该检查关闭时发回的响应，并决定您要采取的行动方案。

    if let httpResponse = response as? NSHTTPURLResponse {
    println("error \(httpResponse.statusCode)")
}

如何使用NSURLSession处理丢失的Internet连接？

2 个答案: