Php image resampling dimension issue

Question

Please i am using the below php image function to resize image when uploading or while displaying the image on page, but i found out that when i tried to resize an original image it doesn't give me the actual dimension that i want.

Example: the original image width is 600 and height is 400 I intend to resize it to 180 X 180, but it gave me width 180 and height 120. Please i don't know what the problem is can anyone help?

<?php
function CroppedThumbnail($imagename, $meta, $newpath, $imgwidth, $imgheight, $rename, $imgQuality){
// Set a maximum height and width
    $width = $imgwidth;
    $height = $imgheight;
    $url = '';
//Get image info

    $info = @getimagesize($imagename);
    $fileparts = pathinfo($imagename);
// Get new dimensions
    $imageAvatar =  substr($fileparts['filename'],0,5) . '-' . $imgwidth . 'x' . $imgheight . '.png';
    if(is_dir($newpath . '/') && file_exists($newpath . '/' . $imageAvatar)){
       return $url . $newpath . '/' . $imageAvatar;
    }else{      
    list($width_orig, $height_orig) = $info;
    $ratio_orig = $width_orig/$height_orig;
    if ($width/$height > $ratio_orig) {
        $width = $height*$ratio_orig;
    } else {
        $height = $width/$ratio_orig;
    }
// Resample
    if ($info['mime'] == 'image/jpeg'){
        $image = imagecreatefromjpeg($url.$imagename);
    }
    else if ($info['mime'] == 'image/gif'){
        $image = imagecreatefromgif($url.$imagename);
    }
    else if ($info['mime'] == 'image/png'){
        $image = imagecreatefrompng($url.$imagename);
    }
    $image_p = imagecreatetruecolor($width, $height);
    imagecopyresampled($image_p, $image, 0, 0, 0, 0, $width, $height, $width_orig, $height_orig);
//Rename the image
    if($rename == true){
        $newName = substr($fileparts['filename'],0,5) . '-' . $imgwidth . 'x' . $imgheight;
        $newFileName = $newpath . '/' . $newName . '.png';
    }else{
        $newFileName = $newpath . '/' . $imagename;
    }
// Output
    imagejpeg($image_p, $newFileName, $imgQuality);
    return $url . $newFileName;
    }
}

Answer 1

This because you are keeping the original aspect ratio of the image.

This is in the following line of code:

$ratio_orig = $width_orig/$height_orig;
if ($width/$height > $ratio_orig) {
    $width = $height*$ratio_orig;
} else {
    $height = $width/$ratio_orig;
}

If you don't want to keep the aspect ration just delete those lines.