使用此时出现语法错误->;db->;CodeIgniter中的escape_str()函数


Syntax error while using this->db->escape_str() function in CodeIgniter

我想通过以下方式将输入数据发送到CodeIgniter应用程序中的SQL查询:

  $sdata['department'] = $this->input->post('department'); //Data as text input 
  $sql = "SELECT MAX(roll) FROM student WHERE department = ".this->db->escape_str($sdata['department']);

上面的代码在我的控制器里。

我不断得到以下错误:

  Parse error: syntax error, unexpected '->' (T_OBJECT_OPERATOR)

最有可能的是,这是一个语法错误,但我只是不知道我的代码的哪一部分应该负责。

我还浏览了CodeIgniter用户指南中的"查询"部分,但没有明确解释。

有人能告诉我我的错误在哪里吗?我想做的正确语法是什么?

我的控制器-

   <?php
   if ( ! defined('BASEPATH')){ exit('No direct script access allowed');}
   class Student extends CI_Controller
   {
       function __construct()
       {
           parent::__construct();
           #$this->load->helper('url');
           $default_roll = '20141000';
           $this->load->model('student_model');
           $this->load->helper('string');
       }
       //Show all Students
       public function index()
       {
            $data['student_list'] = $this->student_model->get_all_students();
            $this->load->view('student_view', $data);
       }

      //Insert a student
      public function insert_student_db()
      {
           $sdata['name'] = $this->input->post('name');
           $sdata['department'] = $this->input->post('department');
           $sql = "SELECT MAX(roll) FROM student WHERE department = ".this->db->escape_str($sdata['department']);
           $query = $this->db->query($sql);
           $rolltext = substr($query, 9);
           $year = substr($query, 3, -3);
          if($rolltext == NULL && $year == NULL)
          {
               $rolltext = str_pad(1,3,'0',STR_PAD_LEFT);
               $year = '2014';
          }
         else
         {
             $rolltext++;
             $rolltext = str_pad($rolltext,3,'0',STR_PAD_LEFT);
             if($rolltext == '100')
             {
                  $rolltext = str_pad(1,3,'0',STR_PAD_LEFT);
                  $year++;
             }
         }  
         $sdata['roll'] = $sdata['department'].$year.$rolltext;
         $sdata['email'] = $this->input->post('email');
         $sdata['mobile'] = $this->input->post('mobile');
         $res = $this->student_model->insert_student($sdata);
        if($res)
        {
            header('location:'.base_url()."index.php/student/".$this->index());
        }
    }   
}
?>  

[由用户Ismael Miguel]提供

问题出在我的控制器student.php:内的这条线路上

  $sql = "SELECT MAX(roll) FROM student WHERE department = ".this->db->escape_str($sdata['department']);

一个非常愚蠢的错误,在一件非常基本的事情上。PHP中的所有变量都必须在前面加一个'$'号。我只是错过了。

所以,在纠正了这一点之后,它看起来是这样的:

$sql = "SELECT MAX(roll) FROM student WHERE department = ".$this->db->escape_str($sdata['department']);

现在运行良好。

PHP无法识别没有"$"符号的this。这就是它抛出解析错误的原因。