データベース接続をPHPでJSON配列を取得私はこのようなJSON配列を取得するために、データベースからの私の結果を連結しようとしています

Question

require "conn.php"; 
$mysql_qry = "SELECT u.* FROM friends f , users u WHERE u.ID = f.FriendID and f.UserID =$ID"; 
$result = mysqli_query($conn ,$mysql_qry); 
if(mysqli_num_rows($result) > 0) { 
    while($row = mysqli_fetch_assoc($result)) { 
     $arr.= array("user" => array(array("ID"=>$row["ID"],"Name"=>$row["Name"],"Email"=>$row["Email"],"Password"=>$row["Password"],"Image"=>$row["Image"],"Profession"=>$row["Profession"],"status" => "1","call" => "login"))); 

    } 
    echo json_encode($arr); 
} 

：

{ 
"user":[ 
    {"ID":"1", "message":"Response code : 200"} 
    {"ID":"2", "message":"Response code : 200"} 
    {"ID":"3", "message":"Response code : 200"} 
    {"ID":"4", "message":"Response code : 200"} 
    {"ID":"5", "message":"Response code : 200"} 
    ]  
} 

は、ユーザーのリストは、クエリによって返さを

Answer 1

適切なコードは：

$arr = array('user' => array()); 
if(mysqli_num_rows($result) > 0) { 
    while($row = mysqli_fetch_assoc($result)) { 
     $arr['user'][] = array( 
      "ID" => $row["ID"], 
      "Name" => $row["Name"], 
      "Email" => $row["Email"], 
      "Password" => $row["Password"], 
      "Image" => $row["Image"], 
      "Profession" => $row["Profession"], 
      "status" => "1", 
      "call" => "login" 
     ); 
    } 
} 
echo json_encode($arr);