mysql - 如何回显mysql选择结果codeigniter

Question

我在 codeigniter 中编写了以下 mysql 查询

  $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;
    }               

回声$结果；它什么都不打印；与 print_r($result); Array ( [count(leave_status)] => 7 ) 这样的数组

但是如何将这个值 7 返回给回显呢？

Answer 1

$result 是一个associative array 你不能直接echo it.Access 的值>leave_status 使用 $result['leave_status']。如下所示。

$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
    }