使用 Swift 2 和 PHP 将 JSON 编码为 Xcode 7

Question

我在 let task = session 行出错，应用程序运行但是当 json 尝试加载应用程序时停止并在控制台中显示致命错误：

unexpectedly found nil while unwrapping an Optional value

我只是想显示我服务器上的 JSON，当我更新到 Xcode 7

时，这个问题就开始了

class ViewController: UIViewController {


override func viewDidLoad() {
    super.viewDidLoad()
    // Do any additional setup after loading the view, typically from a nib.
    
    let jsonUrl = "{http://test.yozzibeens.com/AppTesting/JsonArray.php}"
    
    let session = NSURLSession.sharedSession()
    let shotsUrl = NSURL(string: jsonUrl)
    
    
    let task = session.dataTaskWithURL(shotsUrl!)
        {
        (data, response, error) -> Void in
        
        do {
            let jsonData = try NSJSONSerialization.JSONObjectWithData(data!, options:NSJSONReadingOptions.MutableContainers ) as! NSDictionary
            
            print(String(jsonData["ip"]!))
        } catch _ {
            // Error
        }
    }
    
    task.resume()
    
}
override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

Answer 1

您需要转义花括号：

let jsonUrl = "{http://test.yozzibeens.com/AppTesting/JsonArray.php}"
let escapeJSONURL = jsonUrl.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)!

let shotsUrl = NSURL(string: escapeJSONURL)

使用 Swift 2 和 PHP 将 JSON 编码为 Xcode 7

Encode JSON to Xcode 7 with Swift 2 and PHP

json

ios

swift

xcode7

swift2