用PHP显示图像? [关闭]

I have a webpage that I want to dynamically display content depending on the path of the page. For example, I may want http://www.page.php?input=red or maybe I'll want http://www.page.php?input=blue. In both cases, the actual content is located in ./$dir/.

I've got the gist of this down, but I can't for the life of me figure out images. I have an image in each directory at ./$dir/subdir/images/sample.png. How do I output HTML that dynamically alters the image element?

Here is sample code to demonstrate my question.

<html>
    <?php
        $dir = $_GET["input"];
        $img = imagecreatefrompng("./$dir/subdir/images/sample.png");
    ?>
    <body>
        <img src="???" alt="sample image" />
    </body>
</html>

I want to display ./red/subdir/images/sample.png when I have input=red, ./blue/subdir/images/sample.png when input=blue, etc.

Many thanks in advance for any help.

Edit:
I don't even know if I can store an image into a variable as seen on line 4. I'm really new to PHP.

If only one image:

<img src="<?= $img ?>" alt="sample image" />

If loads of images, use an array.

I would also set the alt text on a database, pull it from there and write it the same way I suggested writing the src.

Just so you can understand:

<?= $img ?>

Does exactly the same thing as

<?php echo $img; ?>

Except with a bit more class.

edit: Typo on the indent made my code disappear.

if (!empty($_GET['input']) {
  $path = './'.$dir.'/subdir/images/sample.png';
  if (file_exists($path)) {
    print '<img src="'.$path.'" alt="sample image" />';
  }
}