提供可变内容的文件下载。 PHP

Question

If we have the following

<form method="post" action="">
<input type="hidden" name="token" value="1">
<input type="submit">
</form>

<?php

if(isset($_POST['token']))
{    

    $v = 'NAME,COLOR
          jack,blue
          ivan,gold';

    header("Content-Type: text/csv");
    header("Content-Disposition: attachment; filename=\"file.csv\"");
    header('Expires: ' . gmdate('D, d M Y H:i:s', gmmktime() - 3600) . ' GMT');
    //header("Content-Length: " . filesize($path));

    echo $v;

}

When I run the script, the browser prompts me to download, when the file is downloaded and I check the contents, it also contains the HTML(above the $v data) markup in the file.

I wish to have a file consisting of just the $v variable data.

Answer 1

Anything you output will be output. An HTTP request is for one resource at a time. You can't expect to send a web page and not expect to have that as part of your file content. PHP cannot read your mind and know what part of your code is for the browser, and what part is for the downloaded resource.

You need to make a separate script that does nothing but return a single resource... the file to be downloaded. Start your script with <?php, no whitespace. If your file is purely PHP (and it probably will be), don't end it with ?> to avoid whitespace at the end.