2017-06-15 46 views
-2

我是fecthing它错了吗?我使用FETCH_ASSOC
它不能正常工作,有人可以帮我吗?我不知道什么是错用它
这是我的代码我在fecthing它错了吗? FETCH_ASSOC

if(!empty($_POST)){ 
     if (isset($_POST['id']) && $_POST['id'] > 0){ 
      $id=$_POST['id']; 
      GetSuppliersView(); 
     } 
    } 
    function GetSuppliersView() { 
     global $db; 
     global $id; 
     try{ 
      $query = $db->query(" 
    SELECT 
     suppliers.*, 
     third_party_services.name as servicename 
    FROM 
     suppliers 
    INNER JOIN 
     suppliers_services 
    ON 
     suppliers.id = suppliers_services.supplier_id 
    LEFT JOIN 
     third_party_services 
    ON 
     suppliers_services.id = third_party_services.id 
    WHERE 
     suppliers_services.supplier_id = $id 
      "); 
      $row=$query->fetch(PDO::FETCH_ASSOC); 
      $result['success'] = true; 
      $result['result'] = $row; 
      echo json_encode($result); 
     return true; 
     } catch (PDOException $pe) { 
      return false; 
     } 
    } 

你能帮助我吗?

+0

哪里了'$ result'是从哪里来的? – Swellar

+0

我觉得你的意思'$结果= $查询 - >取(PDO :: FETCH_ASSOC);'和'不$行= $查询 - >取(PDO :: FETCH_ASSOC);' – Swellar

+1

反过来错误报告,找出 – Ghost

回答

-1

请试试这个

$query->setFetchMode(PDO::FETCH_CLASS, 'Person'); 
$row = $query->fetch(); 

,而不是这个

$row=$query->fetch(PDO::FETCH_ASSOC);