asked on

how can i resize image and keep the aspect ratio ?

Hello here is my script used to upload an image so i need to have the size of the image to 150px
note that i am using easyphp that have php 4.0 so i look for a solution that work on it.
PS: please help by adding the resize image to my code structure listed below

function upload($file,$t,$path)
{
	$post=$_FILES[$file];
	$title=$_POST[$t];
	$rand=rand(100,10000);
	//upload setting
	//$path="../images/products/";	
	//rename
	$remove_these = array(' ','`','"','\'','\\','/');
	//Make the filename unique
	$type1=explode("/" ,$post['type']);
	
		if ( $post['size'] > 0 ) {
		
			if ( $post['type'] != "image/pjpeg" && $post['type'] != "image/gif" && $post['type'] != "image/bmp" && $post['type'] != "image/jpg"  ){
			$picture='1';
			}
			else{
				if ( $post['type'] == "image/pjpeg"){ $picture= $title."-".$rand.'.jpg'; }
				else{ $picture=$title."-".$rand.".".$type1[1];}
					
					if(move_uploaded_file($post["tmp_name"], $path.$picture ) ) { 
						chmod($path.$picture, 0777);
						$my_new_file = $path.$picture;
						} else{ $test="";  }
				}
		
 
		}
		else {
		$picture='0';
		}
return $picture;
}
 
 
//here the trick to render the message to the notification div
$print=upload("image","title","../images/sell/products/");
$script='<script type="text/javascript">';
if ( $print == "0" )
{
$script.='parent.document.getElementById("nimage").innerHTML="KINDLY CHOOSE A VALID IMAGE";'. "\n"; 
$script.='parent.document.getElementById("dimage").style.backgroundColor="#FF4200";'. "\n"; 
}
if ( $print == "1" )
{
$script.='parent.document.getElementById("nimage").innerHTML="NOT AN IMAGE TYPE! | ONLY JPEG, GIF AND BMP IMAGE ACCEPTED";'. "\n"; 
$script.='parent.document.getElementById("dimage").style.backgroundColor="#FF4200";'. "\n"; 
}
else
{
$script.='parent.document.getElementById("nimage").innerHTML="IMAGE SUCCESSFULLY UPLOADED";'. "\n"; 
$script.='parent.document.getElementById("dimage").style.backgroundColor="#ABCDEC";'. "\n"; 
$_SESSION['sellimage']=$print;
 
}
$script.='</script>';
 
echo $script;

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ASKER CERTIFIED SOLUTION

Sander Stad

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Styleminds

ASKER

there many miss type in this script i try to fix but however there still some question mark on this ! where these variable are not delcared anywhere in the function

if($sourceHeight < $sourceWidth){
$longSide = $oldImage.Width;
}

and the function imagecreatefromjpeg() give this error
Fatal error: Call to undefined function: imagecreatefromjpeg() in i:\easyphp1-8\www\audiominds\newmodules\uploadimage.php on line 83

so please advice

Sander Stad

Sorry made a little typo:

if($sourceHeight < $sourceWidth){
$longSide = $sourceWidth;
}

Styleminds

ASKER

this function return that is undefined !
imagecreatefromjpeg($my_new_file);
how can i fix it ?

Sander Stad

Assuming you're on Windows:
Open a text editor (notepad, etc.).
Open php.ini.
Look for a line like this:
;extension=php_gd.dll
Erase the semicolon (;).

Styleminds

ASKER

there is only
extension=php_gd2.dll
and it didn't have any semocolon!
any idea ! how can i get this function to work ! i just need to remind what i advice above i am using easy php with php4 so i need a solution for this version i don't want to upgrade for the time being.
Thanks.

Sander Stad

can you get the phpinfo. do you see anything like GD in there?

<?php
 
phpinfo();
 
?>

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Styleminds

ASKER

sorry i get lost ! where i can find this also in php.ini? does this available in php folder as i found another one in the apashe server
please advise.

Sander Stad

no create a php file and paste the above code in it.
run the file and you'll get all the modules and other information about your current php configuration.