Variable loses its value PHP

Question

I have the code below where my variable $layout loses it's value and I have no idea why, the layoutContent is just a method that get the template of my website, then renderOnlyView get the view supposed to render in that template

$layout = $this->layoutContent($layoutParams);
echo ($layout);
$view = $this->renderOnlyView($view, $params);
echo ($layout);

the whole method where the bug occurred

public function renderView(string $view, array $params = [], array $layoutParams = [])
{
    $layout = $this->layoutContent($layoutParams);
    echo ($layout);
    $view = $this->renderOnlyView($view, $params);
    echo ($layout);

    $layout = str_replace('{{ title }}', $layoutParams['title'] ?? 'No title', $layout);
    
    return str_replace('{{ body }}', $view, $layout);
}

and renderOnlyView

protected function renderOnlyView(string $view, array $params): ?string
{
    foreach ($params as $key => $param) {
        $$key = $param;
    }
    ob_start();
    $viewFilePath = Application::$ROOT_DIR . "/views/$view.blade.php";
    if (file_exists($viewFilePath))
        include_once $viewFilePath;
    else
        echo "the view <b>$view</b> is not found";

    return ob_get_clean();
}

layoutContent just in case

public function layoutContent(array $params = [])
{
    $layout = Application::$APP->controller->layout ?? 'main';
    ob_start();
    include_once Application::$ROOT_DIR . "/views/layout/$layout.layout.php";
    $output = ob_get_clean();
    foreach ($params as $key => $param) {
        $output = str_replace('{{ ' . $key . ' }}', $param, $output);
    }

    return $output;
}

more infos: PHP Version: 7.4.3, system ubuntu 20.04

Answer 1

你错过了一个点，这个函数被调用了两次，第一次是在你渲染我的视图时，第二次是当你视图抛出一个错误并且在捕捉它并尝试显示时再次调用 renderLayout，导致 include_once 返回 null因为它已经包含了 xD

Answer 2

好吧，伙计们发现问题，我错过了一个点，这个函数被调用了两次，第一次是当我渲染我的视图时，第二次当我的视图抛出错误并在捕获它并尝试显示renderLayout再次被调用，导致 include_once 返回 null，因为它已经包含在内：D