iOS Swift JSON檢索數據問題

Question

我想做什麼：檢索我的靜態單元格標籤我的JSON Query的返回值。

我的問題是什麼：我不能這樣做。無法返回單個值來更改我的標籤。

我試過了什麼：首先，我安裝了Alamofire和SwiftyJSON庫。其次，我做了我的PHP腳本，導致正確的JSON數據。我在故事板中創建了標籤，並在我的UITableViewController中創建了@IBOutlet UILabel。第三，我做了我的Alamofire請求，我可以得到我的整個數據的輸出。

我不能做的事：我不能讓個人領域取代我的標籤。我想從我的JSON中檢索[用戶名] [性別] [位置] [生日] [手機] [簽名]，並用此返回來替換我的標籤。但是，當我想檢索[手機]，例如，我沒有結果。

我也覺得我有我的JSON的一個問題，因爲它似乎並沒有返回數組，但只解釋

我tableviewcontroller

  import UIKit 
import Alamofire 
import SwiftyJson 

class PersonalDetails: UITableViewController { 

    required init(coder aDecoder: NSCoder) { 
     println("init PersonalDetails") 
     super.init(coder: aDecoder) 
    } 

    deinit { 
     println("deinit PersonalDetails") 
    } 

    var usersData = [PersonalDetailsData]() 

    @IBOutlet weak var dataUsername: UILabel! 
    @IBOutlet weak var dataGender: UILabel! 
    @IBOutlet weak var dataArea: UILabel! 
    @IBOutlet weak var dataBirthday: UILabel! 
    @IBOutlet weak var dataMobilePhone: UILabel! 
    @IBOutlet weak var dataSignature: UILabel! 

    override func viewDidLoad() { 
     super.viewDidLoad() 

     let prefs:NSUserDefaults = NSUserDefaults.standardUserDefaults() 
     var username = prefs.valueForKey("USERNAME") as NSString 
//load and parse the JSON into an array 
     Alamofire.request(.GET, "http://mywebsite/app/data/jsonpersodata.php", parameters: ["username": username]).responseJSON { (request, response, data, error) in 

      let swiftyJSONObject = JSON(data!) 
      if (error != nil) 
      { 
       // got an error in getting the data, need to handle it 
       println("error calling GET usersdata") 
       println(error) 
      } 
      else if let data: AnyObject = data 
      { 
       // handle the results as JSON, without a bunch of nested if 
       let userdata = JSON(data) 
       if let mobilephone: String = userdata [0]["mobilephone"].stringValue { 
       self.dataMobilePhone.text = mobilephone 
       } 

           } 

     } 

    } 



    override func didReceiveMemoryWarning() { 
     super.didReceiveMemoryWarning() 
      } 

    @IBAction func logoutTapped(sender : UIButton) { 

     let appDomain = NSBundle.mainBundle().bundleIdentifier 
     NSUserDefaults.standardUserDefaults().removePersistentDomainForName(appDomain!) 

     self.performSegueWithIdentifier("logout", sender: self) 
    } 


    } 

我的課

import Foundation 

class PersonalDetailsData: NSObject { 

    var dataUsername:String? 
    var dataGender:String? 
    var dataArea:String? 
    var dataBirthday:String? 
    var dataMobilePhone:String? 
    var dataSignature:String? 
} 

的json

[ 
    { 
    "username": "username1", 
    "gender": "?", 
    "location": "??", 
    "birthday": "1983/01\/16", 
    "mobilephone": "136777777", 
    "signature": null 
    } 
] 

我的PHP

<?php 
header('Content-type: application/json'); 
/* include db.config.php */ 
include_once("config.php"); 
// Get user id 
$id = isset($_GET['username']) ? mysql_real_escape_string($_GET['username']) : 「」; 

if(empty($id)){ 
    $data = array ("result" => 0, "message" => 'Wrong user id'); 
} else { 
// get user data 
$sql = mysql_query("SELECT username, gender, location, birthday, mobilephone, signature FROM users WHERE username='$id'"); 

$data = array(); 

while ($row = mysql_fetch_array($sql, MYSQL_ASSOC)) { 
     $row_array['username'] = $row['username']; 
     $row_array['gender'] = $row['gender']; 
     $row_array['location'] = $row['location']; 
     $row_array['birthday'] = $row['birthday']; 
     $row_array['mobilephone'] = $row['mobilephone']; 
     $row_array['signature'] = $row['signature']; 
     //push the values in the array 
     array_push($data,$row_array); 
} 
echo json_encode($data); 

mysql_close($conn); 
/* JSON Response */ 
} 
?> 

Answer 1

您傳回JSON實際上是一個數組，而不是你期待的對象的直接值（見[和]在你生JSON？他們代表一個數組）。

這裏是你可以做你的JSON數組解析的第一個項目是什麼：

  // handle the results as JSON, without a bunch of nested if 
      let userdata = JSON(data) 
      if let mobilephone: String = userdata[0]["mobilephone"].string { 
       self.dataMobilePhone.text = mobile phone 
      } 
      // etc.