在非对象上调用成员函数 result_array()答案

【问题标题】：Call to a member function result_array() on a non-object在非对象上调用成员函数 result_array()
【发布时间】：2015-02-06 06:42:52
【问题描述】：

请帮忙，我不知道怎么了，这是我的模型

    public function searchstudents(){
    $establishment_name = $this->input->post('name');
    $address = $this->input->post('address');
    $establishment_name2 = "Nicole's Apartment";


        $query = $this->db->query("SELECT u.id as id, u.first_name as firstname, u.middle_name as                  middlename, u.last_name as lastname, e.name as establishmentname FROM users u, beds b, room r, establishment e WHERE  r.establishment_id = e.id AND u.id = b.student_id  AND b.room_id = r.room_id AND e.name = '$establishment_name2' ");

  $this->db->query($query);

    return $query->result_array(); 

}

当我删除 AND e.name = '$establishment_name2' 时，它可以工作，但是当我包含它时，它会导致调用非对象上的成员函数 result_array()。我在mysql中尝试了查询，它可以工作，但在codeigniter中不起作用。

【问题讨论】：

您可以通过将 $query 打印到您的页面来检查查询以及发送到您的数据库的内容:-) 如果问题仍然没有意义。您可以将打印的查询复制/粘贴到您的实际数据库管理系统的 SQL 部分，以便在那里测试运行它。您还可以通过将 sql 字符串粘贴到您喜欢的代码编辑器中来捕获一些格式错误。如果有语法突出显示，您将看到此查询的问题，即单引号之间的撇号处颜色发生变化。

标签： mysql codeigniter

【解决方案1】：

Noooo，看你可以直接这样做

$this->db->query('SELECT u.id as id, u.first_name as firstname, u.middle_name as                  middlename, u.last_name as lastname, e.name as establishmentname FROM users u, beds b, room r, establishment e WHERE  r.establishment_id = e.id AND u.id = b.student_id  AND b.room_id = r.room_id AND e.name = ?', array($establishment_name2));

$this->db->query() 接受 3 个参数：

1 - Query
2 - Array (with parameters, all parameters are clean)
3 - Escape string.

【讨论】：

伙计，看这个页面ellislab.com/codeigniter/user-guide/database/queries.html
看看这个（更有条理）$this->db->select(' u.id as id, u.first_name as firstname, u.middle_name as middlename, u.last_name as lastname, e.name as establishmentname '); $this->db->from('users u, beds b, room r, establishment e'); $this->db->where('r.establishment_id', 'e.id'); $this->db->where('u.id', 'b.student_id'); $this->db->where('b.room_id', 'r.room_id'); $this->db->where('e.name', $establishment_name2);

【解决方案2】：

这是一个刻度问题：

$establishment_name2 = "Nicole's Apartment";

变成：

e.name = 'Nicole's Apartment'

在您的查询中

使用 addlashes() 或像 Mudshark 所说的那样逃跑

【讨论】：

【解决方案3】：

你需要转义$establishment_name2:

$query = $this->db->query("SELECT u.id as id, u.first_name as firstname, u.middle_name as                  middlename, u.last_name as lastname, e.name as establishmentname FROM users u, beds b, room r, establishment e WHERE  r.establishment_id = e.id AND u.id = b.student_id  AND b.room_id = r.room_id AND e.name = ".$this->db->escape($establishment_name2));

【讨论】：