2017-02-28 32 views
1

我笨如何呼應MySQL的選擇結果笨

$query = $this->db->query("SELECT count(`leave_status`) FROM `teacher_attendance` WHERE `teacher_id`='20' AND `leave_status`='6'"); 
    if($query->num_rows() > 0){ 
      $result = $query->row_array(); 
      print_r($result); 
      echo $result; 
    }    

回聲$結果寫了下面的MySQL查詢;它什麼都不打印;帶有print_r($ result)的 ; Array([count(leave_status)] => 7)像這樣的數組

但是如何讓這個值返回7?

回答

0

使用的foreach

$query = $this->db->query("SELECT count(`leave_status`) as count FROM `teacher_attendance` WHERE `teacher_id`='20' AND `leave_status`='6'"); 

if($query->num_rows() > 0){ 
    foreach ($query->result_array() as $row) { 
     echo $row['count']; 
    } 
} 
+0

我嘗試回聲$行向更多信息> leave_status;並且它的任何內容都不是 – shane

+0

在查詢中使用別名。恩。 「SELECT count('leave_status')as count」then「echo $ row-> count;」 – Jhay

+0

但如果我\t print_r($ row);給這個stdClass對象([count('leave_status')] => 7) – shane

2

$resultassociativearray你不能直接echo it.Access使用下面$result['leave_status']。像的leave_status值。

$query = $this->db->query("SELECT count(`leave_status`) AS leave_status FROM `teacher_attendance` WHERE `teacher_id`='20' AND `leave_status`='6'"); 
    if($query->num_rows() > 0){ 
      $result = $query->row_array(); 
      //print_r($result); 
      $total = $result['leave_status']; 
      echo $total ;//prints 7 
    } 
+0

這就是答案 – shane

+0

@shane你有沒有收到它,我希望它可以幫助你。請隨時接受答案,以便其他人可以得到幫助。 –

0

試試這個

$query = $this->db->query("SELECT count(`leave_status`) as count FROM `teacher_attendance` WHERE `teacher_id`='20' AND `leave_status`='6'"); 
if($query->num_rows() > 0){ 
     $result = $query->row_array(); 
     print_r($result); 
     echo $result['count']; 
} 
0

在你的模型:返回$結果

在控制器:$數據= $這個 - > your_model-> your_function();

在視圖:回聲$ leave_status

1

您可以使用行()的迴響單列結果

$query = $this->db->query("SELECT count(`leave_status`) as leaveCount FROM `teacher_attendance` WHERE `teacher_id`='20' AND `leave_status`='6'"); 
if($query->num_rows() > 0){ 
     echo $query->row()->leaveCount; 
} 

參考this link