當我不做var_dump時，php不加載腳本

Question

我用一個函數構建css鏈接，但如果我沒有在最後做一個var_dump，那麼css將無法工作。我錯過了什麼或沒有看到？

代碼：

private function buildCssLinks(){ 
    $files = $this->findFiles(dirname(dirname(__FILE__))."/css","css"); 

    foreach ($files as $id) 
    { 
     $pathInfo = pathinfo($id); 
     $fileName = $pathInfo['basename']; 

     $files[] = '<link rel="stylesheet" type="text/css" href="css/' . $fileName . '">'; 
    } 
    return implode("",$files); 
} 

回報

1st return

，但是當我在我的代碼

private function buildCssLinks(){ 
    $files = $this->findFiles(dirname(dirname(__FILE__))."/css","css"); 

    foreach ($files as $id) 
    { 
     $pathInfo = pathinfo($id); 
     $fileName = $pathInfo['basename']; 

     $files[] = '<link rel="stylesheet" type="text/css" href="css/' . $fileName . '">'; 
    } 
    var_dump($files); 
    return implode("",$files); 
} 

添加一個變種轉儲第二回

2nd return

調用__construct

public function __construct($header, $body, $footer) 
{ 
    $this->header = $header; 
    $this->body  = $body; 
    $this->footer = $footer; 


    $this->buildHeader(); 
    $this->buildBody(); 
    $this->buildFooter(); 

    $js  = $this->buildJsLinks(); 
    $css = $this->buildCssLinks(); 
    $this->header = $css; 
    $this->footer = $js; 
} 

Answer 1

你buildCssLinks正在返回的HTML功能的代碼。但是你必須以某種方式把它放到你的html中。也許

$css = $this->buildCssLinks(); 
$this->header .= $css; 

或更可能需要buildCssLinks第一和傳遞到頭部的構建：

$this->css = $this->buildCssLinks(); 
$this->buildHeader(); // within buildHeader echo your this->css