2016-11-10 71 views
3

我試圖從mysql數據庫編碼json並使用CodeIgniter連接表,但是我想要將與group_by錯誤發生的相同數據組合在一起。哪裏不對? SELECT列表的在CodeIgniter上創建json編碼時無法分組

錯誤警報

表達#1不處於GROUP BY子句和包含非聚集列「project.mytb1.id」,這並不功能取決於由子句中組的列;這是不兼容的sql_mode = only_full_group_by

我的腳本

function get_idCode() { 
    $keyword = $this->uri->segment(3); 
    $data = $this->db->from('mytb1') 
          ->join('mytb2', 'mytb2.id = mytb1.id')  
          ->like('mytb1.id',$keyword) 
          ->group_by('mytb1.id')       
          ->get(); 

    foreach($data->result_array() as $row) 
    { 
     $arr['query'] = $keyword; 
     $arr['suggestions'][] = array(
      'value' =>$row->id, 
      'name' =>$row->name 

     ); 
    } 
    echo json_encode($arr); 
} 

回答

2

試試這個

$data = $this->db->select('*') # added 
        ->from('mytb1') # changed 
        ->join('mytb2', 'mytb2.id = mytb1.id')  
        ->like('mytb1.id',$keyword) 
        ->group_by('mytb1.id')       
        ->order_by('mytb1.id', 'ASC') # added      
        ->get(); 
+1

兄弟,解釋將有助於未來的遊客:) – devpro

+0

@Abdulla Nilam酒店,還是一樣沒有改變 –

+0

'print_r($ data-> result_array())'併發布數據 –