如何將此代碼中的錯誤轉換並修復爲CodeIgniter？

Question

這是怎麼代碼轉換爲笨代碼：

search_hotel： - >這是CI_Model

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

我嘗試，但有錯誤：

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

錯誤：

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 

code： - >這是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

認爲這裏有各種各樣的東西。在您返回$ query-> row（）的模型中，您只會返回一個記錄集行。你應該試試$ query-> result（）來代替。我還建議將你的控制器中的代碼移入模型中。所以，你的模型將開始看起來像：

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，你的控制器簡單變得像：

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

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

希望這應該開始指向你在正確的軌道上。有關創建和使用記錄集的完整信息可以在CI文檔中找到 - http://codeigniter.com/user_guide/database/index.html