Smarty在php函数中

时间:2013-02-13 18:16:07

标签: php smarty

我正在尝试与Smarty一起做一些OOP。当我放,例如

$smarty->display('header.tpl');

在构造函数中,一切正常。但是,当我将此代码放在函数中并在构造中调用函数时,没有任何反应。

是否有解决方案,以便我可以在函数中使用代码然后在函数中调用它?

init.php代码:

class init{
    public function __construct(){

    $smarty = new Smarty();
    $smarty->setTemplateDir('templates');
    $smarty->setCompileDir('templates_c');
    $smarty->setCacheDir('cache');
    $smarty->setConfigDir('configs');
    //$smarty->testInstall();

    $smarty->display('header.tpl');
    $smarty->display('content.tpl');
    $smarty->display('footer.tpl');

    //-----------------------------------
    //-----------------------------------
    //check url
    //-----------------------------------
    //-----------------------------------

    if(isset($_REQUEST['params']) && $_REQUEST['params']!=''){
        $aParams = explode('/', $_REQUEST['params']);
        print_r ($aParams);
        if(isset($aParams[0])) $this->lang = $aParams[0];
        if(isset($aParams[1])) $this->page = $aParams[1];
    }

    if(!$this->lang) $this->lang = 'nl';
    if(!$this->page) $this->page = 'home';

    //-----------------------------------
    //-----------------------------------
    //Functions
    //-----------------------------------
    //-----------------------------------

    $this->buildPage();
    $this->buildHeader_Footer();

}

function buildPage(){
    require_once('modules/' . $this->page . '/' . $this->page . '.php');
    if($this->page == 'home') new home($this->lang, $this->page, $this->action, $this->id, $this->message);
    else if($this->page == 'contact') new contact($this->lang, $this->page, $this->action, $this->id, $this->message);

}

function buildHeader_Footer(){
    $smarty->display('header.tpl');
    $smarty->display('footer.tpl');
}

}

Index.php代码:

require('smarty/libs/Smarty.class.php');

require_once ('modules/init/init.php'); 
$init = new init();

1 个答案:

答案 0 :(得分:0)

更新(问题发生变化后)

您似乎希望在构造函数中创建后可以从所有类方法访问$smarty变量。那是错的。可以使用类中的$this访问类变量。所以你必须写:

$this->smarty-> ...

每当使用它时。


由于发布的代码不完整,我无法确定您的解决方案究竟是什么问题。但是你要做的事情应该有效。

举个例子,我会有这样一个类:

class SimpleView {

    /**
     * Note that smarty is an instance var. This means that var
     * is only available via `$this` in the class scope
     */
    protected $smarty;


    /**
     * Constructor will be called if write 'new SimpleView()'
     */
    public function __construct(){
        // note $this
        $this->smarty = new Smarty();
        $this->smarty->setTemplateDir('templates');
        $this->smarty->setCompileDir('templates_c');
        $this->smarty->setCacheDir('cache');
        $this->smarty->setConfigDir('configs');
    }


    /**
     * The build function is public. It can be called from 
     * outside of the class
     */
    public function build(){
        $this->smarty->display('header.tpl');
        $this->smarty->display('content.tpl');
        $this->smarty->display('footer.tpl');
    }
}

并像这样使用它:

$view = new SimpleView();
$view->build();