json_decode의를 받게되며 PDO fetchAll 후 PDO는

Question

PHP를 가져 오지 않았다

public function addFriend($user) 
    {  
      $sth = $this->db->prepare('SELECT user_id FROM users WHERE user_name = ? '); 
      $sth->execute(array($user->friend)); 
     //Works but then everything else not work 
      //return json_encode($sth->fetch());//NOT work 
      return json_encode($sth->fetchAll()); 

      $count = $sth->rowCount(); 
     if ($count == 1) { 

      // fetch one row (we only have one result) 
      $result = $sth->fetchAll();//NOT WORK  
      //$result = $sth->fetch();//NOT WORK      

$ifexist = $this->db->prepare("SELECT friend_id FROM friends WHERE user_id = ? AND friend_id = ?"); 
     $ifexist->execute(array($user->id, $result->user_id)); 

      $count2 = $ifexist->rowCount(); 
     if ($count2 == 1) { 
     //Friend already exist in friendslist 
     return false; 
     }else{ 

     $token = $this->getToken(50); 
     $uid = $user->id; 
     $fid = $result->user_id; 

     $sth = $this->db->prepare('INSERT INTO friends 
            (user_id, friend_id, token) 
           VALUES (:user_id, :friend_id, :token)'); 

     //we loop 2 times over the bindparams 
     //1st time with above settings ($uid & $fid) and 1 time with settings beneath 
     for($i = 1; $i <= 2; $i++) { 
      $sth->bindParam(':user_id', $uid); 
      $sth->bindParam(':friend_id', $fid); 
      $sth->bindParam(':token', $token); 
      $sth->execute(); 

     //token stay the same 
     $token = $token; 

     //switch clients id($uid) as friends id 
     $uid = $result->user_id; 

     //switch friends id ($fid) as clients id 
     $fid = $user->id; 

    } 

    } 
    } 

} 

jQuery를이 : MVC보기에 내가 jQuery로 바닥 글에서 데이터를 추가 : 나는 MVC 모델에서이

function addFriend() { 

$('#indicator').show(); 
//var checkFriendsName = $('#friendsName').val(); 
//alert(checkFriendsName); 
      var User = new Object(); 
      User.id = uiD;  
      User.friend = $('#friendsName').val(); 
     var userJson = JSON.stringify(User); 

      $.post(BaseUrl+"chat/run", 
      { 
       action: 'insert_friend', 
       user: userJson 
      }, 
      function(data, textStatus) { 
       renderaddFriend(data); 
       $('#indicator').hide(); 
    }, 
      'json' 
       ) 
} 

function renderaddFriend(jsonData) { 
var checkInputName = $('#friendsName').val(); 
    if (jsonData.length == 0) {alert(checkInputName+' not found') }//not work 

$.each(jsonData, function(index, user){  
alert(user.user_id);   
}); 

} 

어떻게 할 수 나는 jquery renderaddFriend 함수에 여전히 json_encode return을 전달합니까?

Answer 1

결과 행이 하나 뿐인 경우 ->fetchAll()은 추가 레코드를 가져 오지 않습니다.

이에 코드를 다시 작성 :

$results = $sth->fetchAll(); 

if (count($results) == 1) { 
    $result = $results[0]; 
    // do stuff 
} 

return json_encode($results);