How to generate image file using this PHP function?

Question

I've been looking for a GD based solution for adding a perspective transformation to an image and eventually found something that seems promising: http://www.jqueryit.com/2010/03/set-perspective-of-image-using-php-gd.html

However, I'm not sure how to actually generate a new image using this function. My approach was this:

function perspective($i,$gradient=0.85,$rightdown=true,$background=0xFFFFFF) {
    $mult=5;
    $w=imagesx($i);
    $h=imagesy($i);
    $image=imagecreatetruecolor($w*$mult,$h*$mult);
    imagecopyresized($image,$i,0,0,0,0,$w*$mult,$h*$mult,$w,$h);
    imagedestroy($i);
    $w*=$mult;
    $h*=$mult;
    $im=imagecreatetruecolor($w,$h);
    $background=imagecolorallocate($im,($background>>16)&0xFF,($background>>8)&0xFF,$background&0xFF);
    imagefill($im,0,0,$background);
    imageantialias($im,true);
    $nh=$h-($h*$gradient);
    for ($x=0; $x<$w; $x++) {
        $ni=(($rightdown) ? $x : $w-$x);
        $p=intval($h-(($ni/$w)*$nh));
        if (($p%2)<>0)
            $p-=1;
        $nx=intval(($p-$h)/2);
        imagecopyresampled($im,$image,$x,0,$x,$nx,1,$p,1,$h-1);
        imageline($im,$x,0,$x,-$nx-1,$background);
        imageline($im,$x,$h-1,$x,$h+$nx,$background);
    }
    imagedestroy($image);
    imagefilter($im,IMG_FILTER_SMOOTH,10);
    $i=imagecreatetruecolor($w/$mult,$h/$mult);
    imageantialias($i,true);
    imagecopyresampled($i,$im,0,0,0,0,$w,$h,$w*$mult,$h*$mult);
    imagedestroy($im);
    return $i;
}

$image = perspective("my_image.jpg");

imagejpeg($image , "my_image_converted.jpg");

And unfortunately, it didn't produce an output. What is it that I'm doing wrong?

Do you actually have code in the //... section or is that what you're asking for help with? — Mr. Llama
– Mr. Llama, Commented Aug 7, 2012 at 15:54
@GigaWatt Exactly, the code is on the page I gave the link to. I Just didn't want to fill up this post with ballast, since the function is already there. — Andrei Oniga
– Andrei Oniga, Commented Aug 7, 2012 at 16:00
I added the function to your post because the problem lies within it (or how you're trying to use it, at least). — WWW
– WWW, Commented Aug 7, 2012 at 16:44

WWW · Accepted Answer · 2012-08-07 15:54:29Z

1

Because you can't pass just a filename when the function requires an image resource. Try:

$image = perspective(imagecreatefromjpeg("my_image.jpg"));

Read through the function you took from your link and look specifically where imagecopyresized() is called.

answered Aug 7, 2012 at 15:54

WWW

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4 Comments

Andrei Oniga Over a year ago

I'm afraid it didn't work, though. I changed the first line to "$image = perspective(imagecreatefromjpeg("home-bg-3.jpg"));" and the second one remained unchanged ("imagejpeg($image , "my_image_converted.jpg");"), but with no effect.

WWW Over a year ago

@AndreiOniga So it wrote out my_image_converted.jpg but that image didn't have the perspective effect applied to it, is that what you're saying?

Andrei Oniga Over a year ago

Actually, for me it's not generating a jpeg file whatsoever, not even without any perspective applied to it.

WWW Over a year ago

@AndreiOniga Time to start adding debug statements and checking server error logs then, I'm afraid.

pdizz · Accepted Answer · 2012-08-07 15:54:07Z

0

If you're just trying to display it you should be able to change the header and echo the image.

header("Content-Type: image/jpg");
echo imagejpeg($image , "my_image_converted.jpg");

The client browser will load this like any other jpg.

answered Aug 7, 2012 at 15:54

pdizz

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1 Comment

Andrei Oniga Over a year ago

I'm trying to actually save it as a standalone file, but even your approach didn't produce an output. Any other ideas?

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