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如何让CI框架支持service层_php实例

时间:2021-07-01 10:21:17 帮助过:6人阅读

大家知道CodeIgniter框架式MVC分层的,通常大家把业务逻辑写到Controller中,而Model只负责和数据库打交道。

但是随着业务越来越复杂,controller越来越臃肿,举一个简单的例子,比如说用户下订单,这必然会有一系列的操作:更新购物车、添加订单记录、会员添加积分等等,且下订单的过程可能在多种场景出现,如果这样的代码放controller中则很臃肿难以复用,如果放model会让持久层和业务层耦合。现在公司的项目就是,很多人将一些业务逻辑写到model中去了,model中又调其它model,也就是业务层和持久层相互耦合。这是极其不合理的,会让model难以维护,且方法难以复用。

是不是可以考虑在controller和model中加一个业务层service,由它来负责业务逻辑,封装好的调用接口可以被controller复用。

这样各层的任务就明确了:
Model(DAO):数据持久层的工作,对数据库的操作都封装在这。
Service : 业务逻辑层,负责业务模块的逻辑应用设计,controller中就可以调用service的接口实现业务逻辑处理,提高了通用的业务逻辑的复用性,设计到具体业务实现会调用Model的接口。
Controller :控制层,负责具体业务流程控制,这里调用service层,将数据返回到视图
View : 负责前端页面展示,与Controller紧密联系。

基于上面描述,实现过程:
(1)让CI能够加载service,service目录放在application下,因为CI系统没有service,则在application/core下新建扩展MY_Service.php

代码如下:
<?php
class MY_Service
{
public function __construct()
{
log_message('debug', "Service Class Initialized");
}
function __get($key)
{
$CI = & get_instance();
return $CI->$key;
}
}

(2)扩展CI_Loader实现,加载service,在application/core下新建MY_Loader.php文件:

代码如下:
<?php
class MY_Loader extends CI_Loader
{
/**
* List of loaded sercices
*
* @var array
* @access protected
*/
protected $_ci_services = array();
/**
* List of paths to load sercices from
*
* @var array
* @access protected
*/
protected $_ci_service_paths = array();
/**
* Constructor
*
* Set the path to the Service files
*/
public function __construct()
{
parent::__construct();
$this->_ci_service_paths = array(APPPATH);
}
/**
* Service Loader
*
* This function lets users load and instantiate classes.
* It is designed to be called from a user's app controllers.
*
* @param string the name of the class
* @param mixed the optional parameters
* @param string an optional object name
* @return void
*/
public function service($service = '', $params = NULL, $object_name = NULL)
{
if(is_array($service))
{
foreach($service as $class)
{
$this->service($class, $params);
}
return;
}
if($service == '' or isset($this->_ci_services[$service])) {
return FALSE;
}
if(! is_null($params) && ! is_array($params)) {
$params = NULL;
}
$subdir = '';
// Is the service in a sub-folder? If so, parse out the filename and path.
if (($last_slash = strrpos($service, '/')) !== FALSE)
{
// The path is in front of the last slash
$subdir = substr($service, 0, $last_slash + 1);
// And the service name behind it
$service = substr($service, $last_slash + 1);
}
foreach($this->_ci_service_paths as $path)
{
$filepath = $path .'service/'.$subdir.$service.'.php';
if ( ! file_exists($filepath))
{
continue;
}
include_once($filepath);
$service = strtolower($service);
if (empty($object_name))
{
$object_name = $service;
}
$service = ucfirst($service);
$CI = &get_instance();
if($params !== NULL)
{
$CI->$object_name = new $service($params);
}
else
{
$CI->$object_name = new $service();
}
$this->_ci_services[] = $object_name;
return;
}
}
}

(3)简单例子实现:
控制器中调用service :

代码如下:
<?php
class User extends CI_Controller
{
public function __construct()
{

parent::__construct();
$this->load->service('user_service');
}
public function login()
{
$name = 'phpddt.com';
$psw = 'password';
print_r($this->user_service->login($name, $psw));
}
}

service中调用model :

代码如下:
<?php
class User_service extends MY_Service
{
public function __construct()
{
parent::__construct();
$this->load->model('user_model');
}
public function login($name, $password)
{
$user = $this->user_model->get_user_by_where($name, $password);
//.....
//.....
//.....
return $user;
}
}

model中你只跟db打交道:

代码如下:
<?php
class User_model extends CI_Model
{
public function __construct()
{
parent::__construct();
}
public function get_user_by_where($name, $password)
{
//$this->db
//......
//......
return array('id' => 1, 'name' => 'mckee');
}
}

基本实现思路就是这样的。

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