无法使用表单移动图像

Question

下面的代码显示由包含产品名称的select标签组成的表单，当我选择特定产品时，它允许编辑其详细信息。我能够更新所有文本数据但无法移动图像文件，它给我一个错误，如未定义索引：图像..请帮我解决这个问题。

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<div class="divD">
<p>Product Model Name</p>
<?php
global $target_path1;
$connection = mysql_connect("localhost", "root", "");
$db = mysql_select_db("abc", $connection);					
if (isset($_GET['submit'])) 
{
  $id = $_GET['did'];
  $m1 = $_GET['model_nm'];
  $m2 = $_GET['price'];
  $m3 = $_GET['des'];
  $target_path1 = "Upload/";
  $target_path1 = $target_path1 . basename( $_FILES['image']['name']); 
  if(move_uploaded_file($_FILES['image']['tmp_name'], $target_path1)) 
	{
		echo "The file ".  basename( $_FILES['image']['name'])." has been uploaded";
	}
$query = mysql_query("update product set model_nm='$m1', price='$m2',des='$m3',img1='$target_path1' where id='$id'");
}

$query = mysql_query("select * from product", $connection);
while ($row = mysql_fetch_array($query)) {?>
	<?php echo "<a href='edit_products.php?update={$row['id']}'>{$row['catg_nm']}--{$row['model_nm']}</a>";
	echo "<br />";
	}?>
</div>

<?php
if (isset($_GET['update'])) {
	$update = $_GET['update'];
	$query1 = mysql_query("select * from product where id=$update");
	while ($row1 = mysql_fetch_array($query1)) {?>
		<form class="form" method="get" enctype="multipart/form-data" action="#">
		<?php	echo "<h2>Update Form</h2>";
		echo "<hr/>";
		echo "<input class='input' type='hidden' name='did' value='{$row1['id']}' />";
		echo "<br />";
		echo "<label>" . "Model Name:" . "</label>" . "<br />";
		echo "<input class='input' type='text' name='model_nm' value='{$row1['model_nm']}' />";
		echo "<br />";
		echo "<label>" . "Price:" . "</label>" . "<br />";
		echo "<input class='input' type='text' name='price' value='{$row1['price']}' />";				

		echo "<br />";
		echo "<label>" . "Configuration:" . "</label>" . "<br />";
		echo "<textarea rows='20' cols='15' name='des'>{$row1['des']}";
		echo "</textarea>";
		echo "<label>" . "Images:" . "</label>" . "<br />";
		echo "<input type='file' name='image' id='image' />";
		echo "<br />";
		echo "<input class='submit' type='submit' name='submit' value='update' />";
		echo "</form>";
		}
	}
if (isset($_GET['submit'])) {
	echo '<div class="form" id="form3"><br><br><br><br><br><br>
	<Span>Data Updated Successfuly......!!</span></div>';
	}
?>
<?php mysql_close($connection);	?>		

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