ต้องการให้สรุปยอดสินค้า โดยแสดงชื่อสินค้าที่ซ้ำเพียง 1 รายการ

<form name="frmSearch" method="post" action="<?=$_SERVER['SCRIPT_NAME'];?>">
                <table width="100%" border="0" align="center">
                  <tr>
                    <th width="161" align="left" valign="middle" >ผู้รับผิดชอบ<select name="lmName">
                        <option value="" selected>--ระบุฝ่าย/สาขา--</option>
                        <?
			$strSQL1 = "SELECT * FROM pgroup ";
			$objQuery1 = mysql_query($strSQL1);
			while($objResuut1 = mysql_fetch_array($objQuery1))
			{
			?>
                        <option value="<?=$objResuut1["id_pgroup"];?>">
                        <?=$objResuut1["name_pgroup"];?>
                        </option>
                        <? } ?>
                      </select>ปีการศึกษา<input name="year" type="text" id="year" value="" size="8" /> <input name="submit" type="submit" value="ค้นหา" />                   </th>
                    </tr>
                </table>
            </form>
<? $year = $_POST['year'];
	$name_pgroup = $_POST['lmName'];	
		$strSQL2 = "SELECT * FROM charges inner join project  where project.status_manager='อนุมัติ' and charges.id_project=project.id_project ";		
		if($_POST["year"] != "")
		{
			$strSQL2 .= " AND project.year like '%".$_POST["year"]."' ";
		}
		if($_POST["lmName"] != "")
		{
			$strSQL2 .= " AND project.owner = '".$_POST["lmName"]."' ";
		}
		if($_POST["year"] == "" and $_POST["lmName"] == ""  )
		{
			$strSQL2 .= " and project.id_project='0'";
		}
	$strSQL2 .= " order by charges.list asc ";	
	$objQuery2 = mysql_query($strSQL2) or die ("Error Query [".$strSQL2."]");
	$a=1;	
	?>
<table width="100%" border="1" align="center" cellpadding="2" cellspacing="0" bordercolor="#e76e6f">
                <tr align="center" bgcolor="#2F97E0" class="style8">
                  <td width="461" height="30" bgcolor="#e76e6f">รายการวัสดุ</td>
                  <td width="234" bgcolor="#e76e6f">จำนวน</td>
                  <td width="143" bgcolor="#e76e6f">งบประมาณ</td>
                  </tr>
                <?
	while($objResult2 = mysql_fetch_array($objQuery2))
	{
	?>
                <tr align="center" valign="top" bgcolor="#FFFFFF" >
                  <td height="20" align="left"><span class="style59"> &nbsp;
                    <?
					$id_project=$objResult2["id_project"];
					 $list=$objResult2["list"];
		$strSQL5 	= "SELECT * FROM material where id_material='$list' ";
			$objQuery5 	= mysql_query($strSQL5) or die ("Error Query [".$strSQL5."]");
			$objResult5 = mysql_fetch_array($objQuery5);
			$name_material =$objResult5["name_material"];
			echo $name_material;				
						?>
                    &nbsp; </span></td>
                  <td height="20" align="left"><div align="center"><?=$objResult2["c_quantity"];?><?=$objResult2["unit"];?></div></td>
                  <td align="right" >
                    <? 
				  $status_plan=$objResult2["status_plan"];
				  $status_manager=$objResult2["status_manager"];
				  $price=$objResult2["price"]; echo number_format($price,2); 				  
				$sumbudget=$sumbudget+$price;				  
				  ?>&nbsp;</td>
                  </tr>
				<? $a++;}	?>
                <tr align="center" valign="top" bgcolor="#FFFFFF" >
                  <td height="20" colspan="2" align="right">งบประมาณที่ใช้&nbsp;&nbsp;</td>
                  <td align="right" valign="middle" ><? echo number_format($sumbudget,2) ; ?>&nbsp;</td>
                  </tr>
                
            </table>