How to Return a File in PHP

How to return a file in PHP

I think you want this:

        $attachment_location = $_SERVER["DOCUMENT_ROOT"] . "/file.zip";
        if (file_exists($attachment_location)) {

            header($_SERVER["SERVER_PROTOCOL"] . " 200 OK");
            header("Cache-Control: public"); // needed for internet explorer
            header("Content-Type: application/zip");
            header("Content-Transfer-Encoding: Binary");
            header("Content-Length:".filesize($attachment_location));
            header("Content-Disposition: attachment; filename=file.zip");
            readfile($attachment_location);
            die();        
        } else {
            die("Error: File not found.");
        } 

PHP rest api how to return a file or download

From your download file URL $request_url.

This is force download source code for server side.

// https://stackoverflow.com/questions/7263923/how-to-force-file-download-with-php

header('Content-Type: application/octet-stream');
header("Content-Transfer-Encoding: Binary"); 
header("Content-disposition: attachment; filename=\"" . basename($request_url) . ".jpg\""); 
readfile($request_url);
exit();// end process to prevent any problems.

This is source code for download working on client side.

$url = 'https://my-domain.tld/rest-api/path.php';// change this to your REST API URL.

// https://stackoverflow.com/questions/6409462/downloading-a-large-file-using-curl

$fileext = pathinfo($url, PATHINFO_EXTENSION);
echo $fileext;
set_time_limit(0);
// For save data to temp file.
$fp = fopen (__DIR__ . '/localfile.tmp', 'w+');

//Here is the file we are downloading, replace spaces with %20
$ch = curl_init(str_replace(" ","%20",$url));
curl_setopt($ch, CURLOPT_TIMEOUT, 50);
// write curl response to file
curl_setopt($ch, CURLOPT_FILE, $fp); 
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HEADER, 1);

// get curl response
$response = curl_exec($ch); 

$header_size = curl_getinfo($ch, CURLINFO_HEADER_SIZE);
$header = substr($response, 0, $header_size);
$body = substr($response, $header_size);

fwrite($fp, $body);

curl_close($ch);
fclose($fp);

// rename temp file.
preg_match('/filename=[\'"]?([\w\d\.\-_]+)[\'"]?/i', $header, $matches);
var_dump($matches);
if (isset($matches[1]) && !is_file(__DIR__ . '/' . $matches[1])) {
    // if target rename file is not exists.
    rename(__DIR__ . '/localfile.tmp', __DIR__ . '/' . $matches[1]);
} else {
    // for debug
    echo 'something wrong!<br>';
    var_dump($matches);
    echo '<br>';
    echo 'File was downloaded into ' . __DIR__ . '/localfile.tmp';
    exit();
}

Return from include file

includeme.php:

$x = 5;
return $x;

main.php:

$myX = require 'includeme.php'; 

also gives the same result

This is one of those little-known features of PHP, but it can be kind of nice for setting up really simple config files.

How to return an included file in PHP without using a return

Your problem is that using include essentially treats your template.php file as a single giant "echo" because there is no value returned inside template.php

You could change the template.php file to return a HTML string, and it would achieve what you want, but this is a nightmare to look at and manage.

return '<div>
<br/>' . $data . '<br/>
</div>';

An alternative, is to "capture" the output of the included file. Its a lot nicer on your template.php file and achieves the same thing. The output buffer functions essentially capture any information "echoed" - you need to be careful using this and the placement of your session_start however as it can catch you off guard depending on how your application is bootstrapped.

This is done by using three ob_ functions the main one being ob_get_contents which will be the contents of template.php in the example below.
http://php.net/manual/en/function.ob-get-contents.php

ob_start();
include $_SERVER['DOCUMENT_ROOT'] . 'template.php';
$html = ob_get_contents();
ob_end_clean();

This was your template files stay nice and clean, and the contents are all captured for returning in your json format.

How to return non blade based view in laravel?

I have custom php file (index.php and not index.blade.php) with HTML & PHP codes. How I can use this file as view in laravel and return it?

I would strongly recommend to use Laravel's Blade Engine and stick to the MVC principle of separating your view from your business logic.

However, if you have to use a custom PHP file instead, then you could do it like this:

class YourController extends Controller
{    
   public function index() 
   {
      $content = require 'path/to/your/index.php'; 
      return $content;
   }   

}

If your path leads to the resource directory you could use the resource_path helper:

$content = require resource_path('views/index.php');

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