Imagick Coding?

ATLien

Hey everyone, I am tooking to write a script that will take a image being uploaded, and make a thumbnail of it. Is Imagick the best way to do this? If so, how do I do it? I have never worked with Imagick before, so I am not sure how to code it in PHP. Does anyone have a script I can look at, a tutorial, or can someone explain it to me? I alreay have the uplaod part, I just need to make a 40 X 40 thumb out of it.

silent

You can use GD... it's included with most installations of PHP these days.

Here's a basic function to resize/make a thumbnail - and it even resamples it to maintain higher quality :
NOTE: This came from a book... not sure which one or else I would give credit where it's due... it's SLIGHTLY modified from the book version :

function resize($fname, $output, $max_width, $max_height, $quality = 75) {

  getimagesize($fname);
  $width = $size[0]; 
  $height = $size[1]; 
  $x_ratio = $max_width / $width; 
  $y_ratio = $max_height / $height; 
  if ( ($width <= $max_width) && ($height <= $max_height) ) { 
    $tn_width = $width; 
    $tn_height = $height; 
  }
  else if (($x_ratio * $height) < $max_height) { 
    $tn_height = ceil($x_ratio * $height); 
    $tn_width = $max_width; 
  } 
  else { 
    $tn_width = ceil($y_ratio * $width); 
    $tn_height = $max_height; 
  } 

  $im = imagecreatefromjpeg($fname);
  $dst = imagecreatetruecolor($tn_width,$tn_height); 
  imagecopyresampled($dst, $im, 0, 0, 0, 0,$tn_width,$tn_height,$width,$height); 
  imagejpeg($dst, $output, $quality);
  imagedestroy($im);
  imagedestroy($dst);
}

Hope it helps you out

ATLien

how would I echo this new image? Like that is the varible for the resized image?

ShawnK

You have to create the array: size before you can use it.

It should read:

 $size = getimagesize($fname); 
  $width = $size[0];  

  $height = $size[1];  

  $x_ratio = $max_width / $width;  

  $y_ratio = $max_height / $height;

ATLien

What? I am alittle confused on what you ment.

ShawnK

My post was a response to silents code he posted.

He called getimagesize($fname) which is the correct syntax but if you wanna return/echo the size later you need to assign it a variable.

silent

Originally posted by ShawnK
You have to create the array: size before you can use it.

It should read:
 $size = getimagesize($fname); 
  $width = $size[0];  

  $height = $size[1];  

  $x_ratio = $max_width / $width;  

  $y_ratio = $max_height / $height;
[/B]

Crap... heheheeh... I guess that got cut off when I cut/pasted it.
Good eye though!