ช่วยดูโค้ดการแสดงฐานข้อมูลหน่อยครับ เลื้อก listmenu แล้วมันไม่แสดงข้อมูลคับ

<?php
$checkError = ini_get('error_reporting');
error_reporting($checkError  ^ E_NOTICE);
include 'connectDB.php';
?>
<script type="text/javascript"> 
function show_data1(){
	document.getElementById("item1").value;
return true;
}
function show_data2(){
	document.getElementById("item2").value;
return true;
}

</script>

<!-----------------------------------------------------------------item1------------------------------------------------------------->
	  <select name="users" id="item1" onChange="show_data1()">
                        <option value="">เลือกวัตถุดิบ</option>
                        <?php
                        
                        $strSQL = "SELECT * FROM raw_material";
                        $objQuery = mysql_query($strSQL) or die("Error Query [" . $strSQL . "]");

                        while ($objResult = mysql_fetch_array($objQuery)) 
						{
                            ?>
                            <option value="<?php echo $objResult["rm_id"] ?>"><?php echo $objResult["rm_name"] ?></option>
							<?php
							}
							?>
      </select></td>
      
<?php //มันไม่แสดงของมูลครับต้องทำไงครับ

 if($objResult["rm_id"] != "")
 {
	$strSQL  = "SELECT * FROM have where rm_id = '$objResult' ";
	mysql_query("SET NAMES UTF8");
	$objQuery = mysql_query($strSQL);
	$i = 0;
		while($objResult2 = mysql_fetch_array($objQuery)){
		$i++;
		?>
			<table border='0' cellpadding='0' cellspacing='0' width='50' align='left' style='float:left;'>
			<tr>
				<td style='width: 255px' valign='center'><input type='text' size='3' name='nutrients1_<?php echo $i; ?>' value='<?php echo $objResult2['nutrients'] ?>' readonly /></td>	
			</tr>
		</table>
		
	  <?php }
	  
		$strSQL1  = "SELECT * FROM raw_material where rm_id = '$objResult' ";
		mysql_query("SET NAMES UTF8");
		$objQuery1 = mysql_query($strSQL1);
			
			while($price = mysql_fetch_array($objQuery1)){
			
				$price['price_raw'];
			}

	  }?> 
	  <br />
<!-----------------------------------------------------------------item2------------------------------------------------------------->
	  
	  <select name="users" id="item2" onChange="show_data2()">
                        <option value="">เลือกวัตถุดิบ</option>
                        <?php
                        
                        $strSQL = "SELECT * FROM raw_material";
                        $objQuery = mysql_query($strSQL) or die("Error Query [" . $strSQL . "]");

                        while ($objResult = mysql_fetch_array($objQuery)) 
						{
                            ?>
                            <option value="<?php echo $objResult["rm_id"] ?>"><?php echo $objResult["rm_name"] ?></option>
							<?php
							}
							?>
      </select></td>
      
<?php //มันไม่แสดงของมูลครับต้องทำไงครับ
 
 if($objResult["rm_id"] != "")
 {
	$strSQL  = "SELECT * FROM have where rm_id = '$objResult' ";
	mysql_query("SET NAMES UTF8");
	$objQuery = mysql_query($strSQL);
	$i = 0;
		while($objResult2 = mysql_fetch_array($objQuery)){
		$i++;
		?>
			<table border='0' cellpadding='0' cellspacing='0' width='50' align='left' style='float:left;'>
			<tr>
				<td style='width: 255px' valign='center'><input type='text' size='3' name='nutrients1_<?php echo $i; ?>' value='<?php echo $objResult2['nutrients'] ?>' readonly /></td>	
			</tr>
		</table>
		
	  <?php }
	  
		$strSQL1  = "SELECT * FROM raw_material where rm_id = '$objResult' ";
		mysql_query("SET NAMES UTF8");
		$objQuery1 = mysql_query($strSQL1);
			
			while($price = mysql_fetch_array($objQuery1)){
			
				$price['price_raw'];
			}

	  }?>