Question

如何将此代码转换为CodeIgniter代码：

search_hotel： - ＆gt;这是CI_Model

return mysql_query("select * from hotel_submits where name LIKE '".$searchterm."'")

我尝试但有错误：

$query = $this->db->order_by("id", "desc")->like('name', '$searchterm')->get('hotel_submits');
        return $query->row();

错误：

A PHP Error was encountered
Severity: Warning
Message: mysql_fetch_assoc() expects parameter 1 to be resource, array given
Filename: admin/tour.php
Line Number: 15


A PHP Error was encountered
Severity: Notice
Message: Undefined variable: data
Filename: admin/tour.php
Line Number: 21

代码： - ＆GT;这是CI_Controller

$searchterm = $this->input->post('search_hotel');
$result = $this->model_tour->search_hotel($searchterm);
while ($row = mysql_fetch_assoc($result)) { //this is line 15
//giving names to the fields
$data = array (
   'name' => $row->name,            
);
}
echo json_encode($data);  //this is line 21

Answer 1

在这里想想各种各样的事情。在您返回$ query-＆gt; row（）的模型中，您只返回一个记录集行。您应该尝试$ query-＆gt; result（）。我还建议将控制器中的代码移动到模型中。所以你的模型会开始看起来像：

function search_hotel($searchterm)
{

$query = $this->db->order_by("id", "desc")->like('name', $searchterm)->get('hotel_submits');

$data = array();

foreach ($query->result() as $row)
{
   data[] = $row->name
}

return $data
}

所以你的控制器很简单就像：

function your_controller() {
//Set your $search_term somehow

    echo json_encode($this->model('your_model_name')->search_hotel($search_term);
}

希望这应该开始指向正确的轨道。有关创建和使用记录集的完整信息，请参阅CI文档 - http://codeigniter.com/user_guide/database/index.html

如何将此代码中的错误转换并修复为CodeIgniter？

1 个答案: