ช่วยดูโค๊ด เกี่ยวกับ เลือกรูปภาพให้หน่อยครับ พอดีผมเขียนในรูป jquery แล้ว มันทำงาน แต่ทำงานไม่ตรงความต้องการ

<meta http-equiv="content-type" content="text/html" charset="utf-8">
<script src="jquery-1.11.0.min.js" type="text/javascript"></script>
<script type="text/javascript">

$(function(){
	$('.PICSQL').click(function(){
		var image = $('.PICSQL').attr('name');
alert(image);
		var url = "GalleryShow1.php";
		var dataSet = "image="+image;
		$.post(url,dataSet,function(data){
			$('#youimage').html(data);
		});
	});
});

</script>

<!--<img src="../BF4 embled.jpg" id="name" >
<input type="button" id="PUSH" value="PUSH">
<span id="mySpan"></span>
-->
<? include('connectserver.php'); 
$qr1 = mysql_query("Select * From gallery Order by GalleryID ASC   ");
echo"<table border=\"0\"  cellspacing=\"1\" cellpadding=\"1\"><tr>";	
$intRows = 0;
while ($rs1 = mysql_fetch_array($qr1)){
	$intRows++;
	echo "<td>";?>
<a class="PICSQL"  name="<?=$rs1['GalleryID'];?>">
<img src="uploads/<?=$rs1['GalleryName'];?>" width="230" height="150" ></a><br>
<?
		echo"</td>";
		if(($intRows)%2==0)
		{
		echo"</tr>";
		}
		else
		{
		echo "<td>";
		}	
	}
	echo"</tr></table>";

?>
<span id="youimage"></span>

<span id="gallery"></span>
<? mysql_close($con);?>

<? include ('connectserver.php');
$STd=$_POST['image'];
$strSQL = "Select * from gallery Where GalleryID = '".$STd."'";
$qr = mysql_query($strSQL);
while($rs = mysql_fetch_array($qr)){
echo "<img src='uploads/$rs[GalleryName]' width='150' height='100'>";
}
mysql_close($con);
?>