Loading an image through PHP

Question

I'm trying to load an image through PHP, but I don't know how. The filename is stored in the database, such as image.jpg

if($_GET['image']){
    // Client requesting image, so retrieve it from DB
    $id = mysql_real_escape_string($_GET['image']);
    $sql = "SELECT * FROM $tbl_name WHERE id = '$id' LIMIT 1";
}

The client needs to request an image like so

http://example.com/index.php?image=1

And then it should return the image, so it can be embedded like this:

<img src="http://example.com/index.php?image=1" alt="" />

Is this possible?

joschua011 · Accepted Answer · 2012-03-31T20:02:27.187

23

$img = 'path/to/image.jpg';
header('Content-Type: image/jpeg');
readfile($img);

just tested it

edited Mar 31 '12 at 20:02

answered Mar 31 '12 at 19:52

joschua011

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Also have a look at [readfile](http://php.net/manual/en/function.readfile.php) it "combines" the `file_get_contents` and `echo` steps. – stewe Mar 31 '12 at 20:06
make sure you don't have any extra space before opening the – luciomonter Jul 24 '18 at 12:17

score 2 · Answer 2 · edited Feb 17 '18 at 05:54

2

You can use the GD library for that. You start by creating a resource using a function like http://php.net/imagecreatefromjpeg. You will need to provide the path as a parameter.

After that, you just output the resource using a function like http://php.net/imagejpeg.

Don't forget to send the content type header, and also to use imagedestroy on the resource.

Update:

Consider this sample:

$im = imagecreatefromjpeg('path/to/image.jpg');
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);

edited Feb 17 '18 at 05:54

dda

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answered Mar 31 '12 at 19:55

mishu

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@safarov maybe it would make it easier to extend if certain modifications are needed.. I did not say it is the only way, I just suggested an option.. – mishu Mar 31 '12 at 20:00

score 0 · Answer 3 · answered Mar 31 '12 at 19:52

I suggest you first make a file called image.php. So you will call image.php?id=1

Have image.php header be the image type. header('Content-Type: image/jpeg');

Then you can use the GDImage library in PHP to load the image, and output it. Or you can read the file and output it. The header() is key.

Loading an image through PHP

3 Answers3

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