zend-framework - 如何使用 zend-framework 应用条件样式表 $this->headLink()->appendStylesheet('css/css.css')

Question

谁能帮我为 IE 的 css 做些什么（对于蓝图 css）

我已经做好了

<?php $headlink = $this->headLink();        
    $headlink->appendStylesheet($this->baseUrl('css/blueprint/screen.css') , 'screen, projection')
        ->appendStylesheet($this->baseUrl('css/blueprint/ie.css'), 'screen, projection', "IE")
        ->appendStylesheet($this->baseUrl('css/blueprint/print.css'));
    echo $headlink;
?>

这段代码也不起作用

<?php $headlink = $this->headLink();        
    $headlink->appendStylesheet($this->baseUrl('css/blueprint/screen.css') , 'screen, projection')
        ->appendStylesheet($this->baseUrl('css/blueprint/ie.css'), 'screen, projection', true)
        ->appendStylesheet($this->baseUrl('css/blueprint/print.css'));
    echo $headlink;
?>

更新：：

它应该看起来像

<!--[if lt IE 8]><link rel="stylesheet" href="css/blueprint/ie.css" type="text/css" media="screen, projection"><![endif]-->

Answer 1

$this->view->headLink()->appendStylesheet('/css/ie.css', 'screen', 'IE');
$this->view->headLink()->appendStylesheet('/css/ie6.css', 'screen', 'IE6'); 

Answer 2

你是这个意思吗

->appendStylesheet($this->baseUrl('/css/ie6.css'), "screen", 'IE 6')

带有IE 6in 的部分可以是任何用于包含 IE 特定样式表的表达式

希望有帮助。