How to pass variable to PHP picture

Question

Motivated by this post https://security.stackexchange.com/questions/32852/risks-of-a-php-image-upload-form I want to display my images by

<?php $pathToPicture = "server/www/images/imagexyz1823014719102714123.png"; ?>

<img src="/resources/php/showImage.php"  >

where showImage.php is simply given by

<?php
header('Content-Type: image/jpeg');
readfile($pathToPicture);
?>

But how can I pass the variable $pathToPicture to showImage.php? I do not want to hard-code $pathToPictue into showImage.php.

Answer 1

将image的路径作为get参数传递给showImage.php脚本，如。

wmctrl

在这里，你可以从<?php $pathToPicture = "server/www/images/imagexyz1823014719102714123.png"; ?> <img src="/resources/php/showImage.php?pathToPicture=<?php echo $pathToPicture;?>" > 数组传递变量：

$_GET

为此，我建议为<?php header('Content-Type: image/jpeg'); readfile($_GET['pathToPicture']); ?> 建议使用base64_encode和base64_decode。也不要像这样公开地暴露图像位置的整个路径。看看下面改进的代码

pathToPicture