Passen Sie die JSON-Antwort in Array und Objekte

Question

Ich arbeite an den Web-Service und JSON-Antwort.

Ich habe diese Antwort von der API:

[ 
    [ 
     { 
      "id":"1", 
      "unique_id":"579992ffd1", 
      "contact_name":"qweryyy", 
      "user_phone_number":"03331524145", 
      "created_at":"2016-08-01 15:53:59" 
     }, 
     { 
      "id":"2", 
      "unique_id":"579992ffd1", 
      "contact_name":"qwer", 
      "user_phone_number":"03331524231", 
      "created_at":"2016-08-01 16:04:59" 
     }, 
     { 
      "id":"3", 
      "unique_id":"579992ffd1", 
      "contact_name":"qwer", 
      "user_phone_number":"0333152111", 
      "created_at":"2016-08-01 16:05:08" 
     } 
    ] 
] 

aber ich brauche die Antwort wie diese JSON zeigen:

{ 
    "unique_id":"579992ffd1", 
    "user":[ 
     { 
      "id":"1", 
      "contact_name":"qweryyy", 
      "user_phone_number":"03331524145", 
      "created_at":"2016-08-01 15:53:59" 
     }, 
     { 
      "id":"2", 
      "contact_name":"qweryyy", 
      "user_phone_number":"03331524145", 
      "created_at":"2016-08-01 15:53:59" 
     }, 
     { 
      "id":"3", 
      "contact_name":"qweryyy", 
      "user_phone_number":"03331524145", 
      "created_at":"2016-08-01 15:53:59" 
     } 
    ] 
} 

Hier ist mein Code den Benutzer zu erhalten Array und geben Sie es zurück, um die JSON-Antwort anzuzeigen:

if ($result) { 
     $this->conn = new PDO("mysql:host=$servername;dbname=$dbname", 
$username, $password); 
     $this->conn->setAttribute(PDO::ATTR_ERRMODE, 
PDO::ERRMODE_EXCEPTION); 
     $stmt = $this->conn->prepare("SELECT * FROM contact WHERE 
user_phone_number = '$user_phone_number' "); 
     //$stmt->bind_param("s", $phone_number); 
     $stmt->execute(); 
     $user = $stmt->setFetchMode(PDO::FETCH_ASSOC); 
     $user = $stmt->fetchAll(); 
     //print_r($user); 
     return $user; 

    } else { 
     return false; 
    } 

SHow json Antwort:

if ($user) { 
    $string = ''; 
    $cart = array(); 
    $response["status"] = TRUE; 
    for($x=0;$x<count($user);$x++) 
    { 
     $string = $user[$x]; 
     array_push($cart, $string); 

    } 

     echo json_encode(array($cart)); 


} 

Jede Hilfe die erforderliche Leistung zeigen ??

Answer 1

$query = 'SELECT * FROM contact WHERE user_phone_number = "'.$user_phone_number.'"'; 
$result = mysql_query($query) or trigger_error($query.'<br>'.mysql_error(),E_USER_ERROR); 
$list = Array(); 
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) 
{ 
    $list[] = $row; 
} 
$cart = Array(
    'unique_id' => '579992ffd1'; 
    'user' => $list; 
); 
echo json_encode($cart);