无法遍历此字典并将其组织成自定义结构

Question

我有这段代码用于迭代数组playerForm。

playerForm包含阵列中每个玩家的数组数量，并包含每个玩家的表单。

即

["playerForm": { 
1 =  (
      { 
     date = "2017-01-31"; 
     name = Dicky; 
     result = L; 
     "results_id" = 42; 
    }, 
      { 
     date = "2017-01-26"; 
     name = Dicky; 
     result = L; 
     "results_id" = 41; 
    } 
); 
2 =  (
      { 
     date = "2017-01-25"; 
     name = G; 
     result = W; 
     "results_id" = 38; 
    }, 
      { 
     date = "2017-01-25"; 
     name = G; 
     result = D; 
     "results_id" = 40; 
    } 
3 =  (
      { 
     date = "2017-01-31"; 
     name = Sultan; 
     result = W; 
     "results_id" = 42; 
    }, 
      { 
     date = "2017-01-26"; 
     name = Sultan; 
     result = W; 
     "results_id" = 41; 
    } 
    ); 
    }] 

这是我想使用的代码：

let json = try JSONSerialization.jsonObject(with: data!, options: .allowFragments) as? [String:AnyObject] 
        print (json!) 

        if let dict = json?["playerForm"] as? [String:Any] { 

         print ("step 1") 

         for value in dict { 
          if let arr = value as? [[String:Any]] { 
           print(arr) 
           self.leagueForm = arr.flatMap { Form($0) } 

           for form in self.leagueForm { 
            self.formGuide.append(form.player_result!) 
           } 
           print ("break") 

          } 
         } 


         print (self.formGuide) 
        } 

这是我自定义的结构来组织数据;

struct Form { 
    var player_result: String? 
    var player_name: String? 
    var result_date: String? 
    var result_id: String? 

    init(_ dictionary: [String : Any]) { 
     self.player_result = dictionary["result"] as? String ?? "" 
     self.player_name = dictionary["name"] as? String ?? "" 
     result_date = dictionary["date"] as? String ?? "" 
     result_id = String(dictionary["results_id"] as? Int ?? 0) 


    } 
} 

var leagueForm = [Form]() 

但是，我收到警告：Cast from '(key: String, value: AnyObject)' to unrelated type [[String : Any]] always fails。

这是我的PHP脚本提供初始阵列：

$noPlayers = count($communityPlayersIds); // 
$playerForm = array(); 
$playerForm = $dao->getCommunityForm($communityId, $noPlayers, $communityPlayersIds); 

public function getCommunityForm($communityId, $noPlayers, $communityPlayersIds){ 
$sql = " SELECT IF(player1_id=?, player1_result, player2_result) AS result, IF(player1_id=?, player1_name, player2_name) AS name, date, results_id FROM `results` WHERE (player1_id=? OR player2_id=?) AND community_id=? ORDER BY date DESC Limit 8"; 
$stmt = $this->conn->prepare($sql); 
$i = 0; 
foreach ($communityPlayersIds as $cPI) { 
    $i++; 
    $stmt->bind_param("iiiii", $cPI, $cPI, $cPI, $cPI, $communityId); 
    $stmt->execute(); 
    if ($result = $stmt->get_result()) { 
     while($row = $result->fetch_array(MYSQLI_ASSOC)){ 
      $returnValue[$i][] = $row; 
      } 
     } 
    }return $returnValue; 
} 

echo json_encode (array('playerForm' => $playerForm)); 

所以这个1,2,3在上面的例子是我怎么也收录从MySQL返回的结果。我这样做是为了分离数据，这样我就可以在Swift中构建它了。

我在哪里出错了？

Answer 1

你没有正确解开你的JSON。 json["playerForm"]是一个字典数组，而不是字典。试试这个：

let json = try! JSONSerialization.jsonObject(with: data, options: [.allowFragments]) as! [String: Any] 
var leagueForm = [Form]() 

if let forms = json["playerForm"] as? [[String: Any]] { 
    leagueForm = forms.flatMap { Form($0) } 
} 

---

我认为这是你的样品JSON：

{ 
    "playerForm": [ 
     {"result": "L", "name": "Dicky", "date": "2017-02-10", "results_id": 48}, 
     {"result": "L", "name": "Dicky", "date": "2017-02-10", "results_id": 47}, 
     {"result": "L", "name": "Dicky", "date": "2017-02-09", "results_id": 44}, 
     {"result": "L", "name": "Dicky", "date": "2017-01-16", "results_id": 32}, 
     {"result": "D", "name": "Dicky", "date": "2016-12‌​-12", "results_id": 4}, 
     {"result": "W", "name": "Dicky", "date": "2016-12-12", "results_id": 6}, 
     {"result": "W", "name": "Dicky", "date": "2016-12-12", "results_id": 8} 
    ] 
} 

Answer 2

当您使用value in dict swift返回两条信息：一个键和一个值。因此，一般比较公认的办法是做这样的事情：

for (key, value) in dict ... 

如果预测，对于每个字典项的值将是[[String: Any]]类型的话，我建议做这样的事情：

for (_, value) in dict { 
    if let dictValue = value as? [[String:Any]] { 
     //do the rest with this value 
    } 

} 

请注意，您可以为关键变量名称加上下划线，因为您不打算使用它，因此您不需要将其存储在内存中。

当然，这并不能保证你的代码现在可以工作，但它应该解决这个特定的问题。

Answer 3

我觉得你的问题是在这行代码

if let arr = value as? [[String:Any]] { 

看你的JSON文件，我可以看到你的JSON对象的格式如下：

["playerForm": { 
1 =  (
      { 
     date = "2017-01-31"; 
     name = Dicky; 
     result = L; 
     "results_id" = 42; 
    }, 
      { 
     date = "2017-01-26"; 
     name = Dicky; 
     result = L; 
     "results_id" = 41; 
    } 
);... 

所以，你的JSON根对象是一个[String：Any]字典，你正在做的是正确的。字典只有一个（键，值）对，键为'PlayerForm'。之后，你假设该对的值是[[String：Any]]的类型，这将是一个字典数组数组，我认为它不是:) 我认为这部分：

2 =  (
      { 

是从JSON文件有点故障，我不太清楚，以什么“2 =”应常备的，但它不是一个数组，数组不包含关键，像“ 1,2,3 ......“。

我建议铸造值别的东西，也许尝试

东西=价值？ [字符串：任何]

如果这样的话，那么你可以将该字典中的值转换为NSArray进行迭代。

刚玩的铸有一段时间了:)