ช่วยแนะนำวิธีการเช็คนามสกุลภาพหน่อยครับ พอดีแบบที่ทำอยู่ไม่รู้ว่าถูกหรือไม่ครับ?

 /* File Checking */ 
if (filesize($_FILES['pic']['size']) > 5242880) 
die('Too Big, Maximum Size is <strong> 5MB </strong>. <p> Please Wait As You Are Sent Back To The Change Photo Page... <meta http-equiv="refresh" content="3;URL=java_script_: history.go(-1)">'); 

if (mime_content_type($_FILES['pic']['type']) == "image/gif"); 
elseif (mime_content_type($_FILES['pic']['type']) == "image/jpg"); 
elseif (mime_content_type($_FILES['pic']['type']) == "image/jpeg"); 
elseif (mime_content_type($_FILES['pic']['type']) == "image/bmp"); 
elseif (mime_content_type($_FILES['pic']['type']) == "image/png"); 
else 
die ('<stong> Wrong </strong> Photo File Type, Must be: <strong> <p> Gif <p> Jpg <p> Jpeg <p> Bmp <p> Png </strong> <p> Please Wait As You Are Sent Back To The Change Photo Page... <meta http-equiv="refresh" content="4;URL=java_script_: history.go(-1)">'); 
/* File Checked */ 

<?php 
// $_FILES["file"]["type"] กำหนดนามสกุล
// $_FILES["file"]["size"] กำหนดขนาด
// $_FILES["file"]["error"] ตรวจสอบ Error

if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 20000))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Error: " . $_FILES["file"]["error"] . "<br>";
    }
  else
    {
    // อัพโหลดรูป
    copy($_FILES['img']['tmp_name'], "../image/".$_FILES["img"]["name"])

    echo "Upload: " . $_FILES["file"]["name"] . "<br>";
    echo "Type: " . $_FILES["file"]["type"] . "<br>";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br>";
    echo "Stored in: " . $_FILES["file"]["tmp_name"];
    }
  }
else
  {
  echo "Invalid file";
  } ?>

if(($filetype == "image/jpg")){echo "A";}else{echo "B";}