get an image with an unknown file type

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mattwidge

I'm making a web-app for Conky that's like a Conky script database.

My problem is: I am putting in a preview image in the sidebar.

It will be named like: uploads/script id.gif/jpg/bmp/png

How do I do this?

Jerail

This is just a quick suggestion. There might be something better or more efficient out there that you could do.

First, I'd get a list of all the files in the directory, and then search through it for any image file extensions:

<?php
$files_list = "";
if ($handle = opendir('.')) {
    while (false !== ($file = readdir($handle))) {
        if ($file != "." && $file != "..") {
            $files_list .= $file . ",";
        }
    }
    closedir($handle);
}
?>

Not sure if this is exactly what you were looking for.

mattwidge

How would I implement this?

Mindzai

If I understand the issue, you want to reference a file, you will know the filename, but the extension could be one of a few possibilities. Rather than getting a list of every file, I would be inclined to do it the other way and check the list of extensions. This will presumably be a much shorter list:

<?php

$exts = array('png', 'gif', 'jpg', 'jpeg');
$file = 'images/hello';

$src = '';
foreach ($exts as $ext) {
    if (file_exists("$file.$ext")) {
        $src = "$file.$ext";
        break;
    }
}

printf('<img src="&#37;s" />', $src);

?>

mattwidge

Didn't work.

Here's some more info:
the filename is $id & the filetype will vary. I'm making a thing that uploads an image & is displayed.

Mindzai

Could you be a little more specific than "didn't work?". I tested this on my server and it worked fine.

mattwidge

It didn't show the image.

Jerail

I think you'll need to give us some more specific information about what you want to happen.
The script Mindzai posted will print off links to any images in the images/ directory with the name hello.png, hello.gif, hello.jpg or hello.jpeg. That could be modified slightly to fit what you said originally like this:

<?php

$exts = array('png', 'gif', 'jpg', 'jpeg');
$file = "uploads/$script_id";         // <-- You'd have to define $script_id

$src = '';
foreach ($exts as $ext) {
    if (file_exists("$file.$ext")) {
        $src = "$file.$ext";
        break;
    }
}

printf('<img src="%s" />', $src);

?>

Mindzai

mattwidge;1060373 wrote:
It didn't show the image.

If you can't be bothered to post more than a few words, at least detailing some things like the actual output of the script and any debugging you've tried, I can't be bothered to help you fix it. Maybe if you show a bit of courtesy to the people going out of their way to help you, you will find people more willing to keep on helping.