[英]Load view error codeigniter
I'm new in codeigniter. 我是Codeigniter的新手。 I want to save something database.
我想保存一些数据库。 I have link localhost/index.php/blog/gerilim_controller/40 I want to save 40 to database.
我有链接localhost / index.php / blog / gerilim_controller / 40我想将40保存到数据库。
I wrote some code. 我写了一些代码。 This is my model code: Akim_model.php;
这是我的模型代码:Akim_model.php;
<?php
class Akim_model extends CI_Model
{
function __construct()
{
parent::__construct();
$this->load->database();
}
function ekle($data)
{
$ekle=$this->db->insert('gerilim',$data);
if($ekle){
return 1;
}else{
return 0;
}
}
}
?>
This is my controllers code: Blog.php; 这是我的控制器代码:Blog.php;
<?php
class Blog extends CI_Controller {
function __construct()
{
parent::__construct();
}
public function index()
{
$this->load->helper('url'); }
public function gerilim_controller( $gerilim_id = NULL ) {
echo $gerilim_id;
$data=array('gerilim'=>$this->input->post('gerilim'));
$this->load->model('akim_model');
$sonuc=$this->akim_model->ekle($data);
if($sonuc==1)
echo "Successed";
else
echo "Failed";
}
}
?>
But when $this->load->model('akim_model');
但是当
$this->load->model('akim_model');
its gives an error. 它给出了一个错误。 Where I am wrong?
我哪里错了?
THe file name must match the class name for your model: 文件名必须与您的模型的类名匹配:
<?php
class Akim_model extends CI_Model {
function __construct() {
parent::__construct();
$this->load->database();
}
function ekle($data) {
$ekle=$this->db->insert('gerilim',$data);
if($ekle) {
return 1;
}
else{
return 0;
}
}
}
?>
Also put your conditional statement in curly braces: 还要将条件语句放在花括号中:
if($sonuc==1) {
echo "Successed";
}
else {
echo "Not successed";
}
You should call your model using the class name: 您应该使用类名称来调用模型:
$this->akim_model->ekle($data);
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