2017-03-07 45 views
-2

我需要在Swift 3中通过PHP从mySQL获取我的GPS位置。我试图编写获取数据的代码,但它仍然不起作用,您能否告诉我?如何解析Swift 3中的JSON数据?

JSON从PHP数据:

[{ “ID”: “3752”, “纬度”: “11.2222”, “经度”: “111.2222”, “速度”: “0.000000”,”伏 “:” 3.97" , “百分比”: “87.000000”, “日期”: “2017年3月7日22时53分32秒”}]

夫特3代码:

import UIKit 
//-------- import google map library --------// 
import GoogleMaps 
import GooglePlaces 
class ViewController: UIViewController , GMSMapViewDelegate { 
    var placesClient: GMSPlacesClient! 
    override func viewDidLoad() { 
     super.viewDidLoad()  
     var abc : String = String()   
     //-------- Google key for ios --------// 
     GMSServices.provideAPIKey("XXXXXXXXXX") 
     GMSPlacesClient.provideAPIKey("XXXXXXXXX") 
     //--------set URL --------// 
     let myUrl = URL(string: "http://www.myweb/service.php");   
     var request = URLRequest(url:myUrl!)   
     request.httpMethod = "POST"   
     let postString = "";   
     request.httpBody = postString.data(using: String.Encoding.utf8);   
     let task = URLSession.shared.dataTask(with: request) { (data: Data?, response: URLResponse?, error: Error?) in 
      if error != nil 
      { 
       print("error=\(error)") 
       return 
      }    
      // You can print out response object 
      print("response = \(response)")   

      do { 
       let json = try JSONSerialization.jsonObject(with: data!, options: .mutableContainers) as? NSDictionary 

       if let parseJSON = json {      
        // Now we can access value of latiutde 
        let latitude= parseJSON["latitude"] as? String //<---- Here , which i need latitude value 
        print("latitude = \(latitude)")  
       } 
      } catch { 
       print(error) 
      } 
     } 
     task.resume()  

} 

我试图编写代码,但它显示调试输出上的错误

let responseString = String(data: data, encoding: .utf8) 
    let str = String(data: data, encoding: .utf8) 
    let data2 = str?.data(using: String.Encoding.utf8, allowLossyConversion: false)! 

    do { 
     let json = try JSONSerialization.jsonObject(with: data2!, options: []) as! [String: AnyObject] 
     if let names = json["latitude"] as? [String] { 
      print(names) 
     } 
    } catch let error as NSError { 
     print("Failed to load: \(error.localizedDescription)") 
    }    

} 

错误消息

无法投型的值 '__NSSingleObjectArrayI'(0x1065fad60)至 '的NSDictionary'(0x1065fb288)。

+0

定义“不起作用”。 – rmaddy

+0

另外:http://stackoverflow.com/q/38155436/2415822 – JAL

+1

a l a m o f i r e –

回答

0

尝试将json对象直接转换为Swift表示,以便更底层数据的“Swifty”访问。所以,你不必大惊小怪周围的NSNumber等

guard let json = JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [[String: String]] else { return } 
guard json.count > 0 else { return } 
guard let lattitude = json[0]["lattitude"] else { return } 
print("Lattitude received: \(lattitude)") 

如果你不知道你有一个[String: String]对象数组,你可以用中投[String: Any]替换它,那么你需要do是在读取格子时用可选的cast来检查类型。您可以添加一个链接选项,然后检查isEmpty以检查其是否需要或者出错。

我也建议几乎不会在代码中使用!,尝试更多地依赖可选的链接和保护语句。

Guard statement introduction

注:单行后卫语句不是很详细,可能使其很难调试应用程序。考虑抛出错误或在guard语句的主体中进行更多的调试打印。