如何获取php文件的html生成内容而不是包含该文件

Question

I want to make a cms which generates html pages and creates files to store each page.

Ideally i need something like that:

<?php
$file1 = get_produced_html_from('mainPage.php');
/* write the file to a directory*/


$file2 = get_produced_html_from('ProductsPage.php');
/* write the file to a directory*/
?>

Is there any function i missed instead of require, include, require_once, include_onse etc?

To clarify: I don't need the php code inside the .php file. I need the html content only which means that the php file should be executed first.

Do you believe the solution is something like using http://php.net/manual/en/function.file-get-contents.php by reading http://domain.com/templates/mainPage.php as a html stream?

Thank you a lot.

Answer 1

You need to catch the output from the buffer.

Here is a piece of code I wrote for somebody to demonstrate a very simple view renderer class.

public function render($file) {
    $file = $this->viewPath = APP_ROOT . 'View' . DS . $file . '.php';
    if (is_file($file)) {
        ob_start();
        extract($this->_vars);
        include($file);
        $content = ob_get_contents();
        ob_end_clean();
    } else {
        throw new RuntimeException(sprintf('Cant find view file %s!', $file));
    }

    return $content;
}

It turns on the output buffer (ob_start()) executes the php file and sets the variables and then gets the buffer (ob_get_contents()) and then it cleans the buffer for the next operation (ob_end_clean()). You can also use ob_end_flush() to directly clean and send the buffer. I would not do that and instead do a proper shutdown process of the app and ensure everything is done and was done right without errors before sending the page to the client.

I think I'm going to make the whole code available on Github soon. I'll update the answer then.