Displaying php generated image on a page

Question

I have a function i use in modifying my images, and i want to display thos images on my page alongside other page content.

PHP Image Function :

<?php

 function dynamicImage($nw, $nh, $source, $stype, $dest) {
        $size = getimagesize($source);
        $w = $size[0];
        $h = $size[1];
        switch($stype) {
            case 'gif':
            $simg = imagecreatefromgif($source);
            break;
            case 'jpg':
            $simg = imagecreatefromjpeg($source);
            break;
            case 'png':
            $simg = imagecreatefrompng($source);
            break;
        }
        $dimg = imagecreatetruecolor($nw, $nh);
        $wm = $w/$nw;
        $hm = $h/$nh;
        $h_height = $nh/2;
        $w_height = $nw/2;
        if($w> $h) {
            $adjusted_width = $w / $hm;
            $half_width = $adjusted_width / 2;
            $int_width = $half_width - $w_height;
            imagecopyresampled($dimg,$simg,-$int_width,0,0,0,$adjusted_width,$nh,$w,$h);
        } elseif(($w <$h) || ($w == $h)) {
            $adjusted_height = $h / $wm;
            $half_height = $adjusted_height / 2;
            $int_height = $half_height - $h_height;
            imagecopyresampled($dimg,$simg,0,-$int_height,0,0,$nw,$adjusted_height,$w,$h);
        } else {
            imagecopyresampled($dimg,$simg,0,0,0,0,$nw,$nh,$w,$h);
        }
        header( "Content-type: image/jpeg" );
        imagejpeg( $dimg );
        imagedestroy( $dimg );
        //imagejpeg($dimg,$dest,100);
    }
?>

on php page where i want it displayed ( i call the dynamicImage function ) :

<img id="profilepic" src="<?php $size=getimagesize($pic); $imgx=240; $imgy=($size[1]*$imgx)/$size[0]; dynamicImage($imgx, $imgy, $pic, substr($pic, -3), ''); ?>"  />

but it ends up giving out a blank page.

How do i go about it witout using base encode since i need to support a wide range of older browsers.

Thanks in advance (this site has been so helpful smiles).

Answer 1

You are puting image data to src atribute, if you want to put inline image you must put it in base64.

imagejpeg($dimg, 'tmp_image');
$handle = fopen($filename, "r");
$imgbinary = fread(fopen($imgfile, "r"), filesize($imgfile));

return 'src="data:image/jpeg;base64," . base64_encode($imgbinary) . '"';

or put your function into separete script and change

dynamicImage($imgx, $imgy, $pic, substr($pic, -3), '')

to

"src=\"dynamicimage.php?w={$imgx}&h={$imgy}&fname={$pic}&type=" . substr($pic, -3) . '";

Answer 2

Don't call the php code directly in your image tag

lets say you create a script for generating an image and name it

/myimage.php

code:

<?php

$type = $_GET['type'] || 'png';
// remember to test if variables exist and handle gracefully
$mimetypes = array(
    'jpg' => 'image/jpeg',
    'png' => 'image/png',
    // ad nauseum
);

header('Content-type: '.$mimetypes[$type]);

// handle parsing and output of image here.   

you then use it in your page like this:

<img src="/myimage.php?width=240&height=240&type=jpg" />

Answer 3

Your image needs to be in a separate PHP page to the rest of the page content.

ie your main PHP page would produce the HTML output as normal. That HTML would contain an tag as normal, but the src attribute would contain a link to another PHP program. This is where you went wrong.

So your tag needs to look something like this:

<img src='mypicture.php'>

And then you need a separate PHP program called mypicture.php which produces the image.

Note that if you're producing images from a php program, you need to be careful to only output the image data and nothing else (watch out for blank lines at the end of your code after the ?>). You also need to set the mime type in the headers to image/jpeg (or whatever the format is).