将图像输出到浏览器的函数返回text / html类型

Question

I have the below function to output an image to the browser

function sendImage(string $file, int $browser_cache = 2592000) {

  if (file_exists($file)) {

    $mime = getimagesize($file)['mime'];

    header("Content-Type: ". $mime);
    header("Cache-Control: private, max-age=".$browser_cache);
    header('Expires: '.gmdate('D, d M Y H:i:s', time()+$browser_cache).' GMT');
    header('Content-Length: '.filesize($file));
    readfile($file);
    exit;
  }
}

call the function

$file = $_SERVER['DOCUMENT_ROOT'] . '/images/image.png';
sendImage($file);

The content type output shows text/html and there is no response content. No errors either. The file definitely exists. I cant quite figure out the error in the function.

Answer 1

Your code seems to work as intended. You think you should add a check (e.g. function returns) to avoid possible warnings, and get a fallback if the file doesn't exists.

function sendImage(string $file, int $browser_cache = 2592000) {

    if (!file_exists($file)) return false ;

    $mime = getimagesize($file)['mime'];

    header("Content-Type: ". $mime);
    header("Cache-Control: private, max-age=".$browser_cache);
    header('Expires: '.gmdate('D, d M Y H:i:s', time()+$browser_cache).' GMT');
    header('Content-Length: '.filesize($file));
    readfile($file);
    exit;
}

Then you could check this :

$file = $_SERVER['DOCUMENT_ROOT'] . '/images/image.png';
if (!sendImage($file)) {
    echo "file not found" ;
}

Additionally, if the $file is not an image, getimagesize() will returns false. And getimagesize($file)['mime'] will emit a warning.

$infos = getimagesize($file) ;
if (!$infos) return false ; // Or something else to specialize the fallback.
$mime = $infos['mime'];