ขอคำแนะนำเกี่ยวกับการดึงข้อมูลจากฐานข้อมูลมาแสดงใน textbox แต่แสดงได้ไม่ครบ

<?php

$name = '<center><span style="font-size:9px;font-style:italic;"background-color:#119922;color:#993311;"></center><br />';
$id_product = $_REQUEST['id_product'];
$conn = mysql_connect("localhost","root","mysql");
$query = "use project"; //ไม่เข้าใจบรรทัดนี้อ่ะ
$result = mysql_query($query) or die ("Query failed");

$query = "select * from product where id_product = '$id_product'";

$result = mysql_query($query) or die ("Query failed");
$row = mysql_fetch_array($result);
?>
<form action="up_product.php?id_product=<?=$id_product?>" method="post">
Edit product<br />
รหัสสินค้า : <input type="text" name="id_product" value="<?=$row[id_product]?>" /><br />
.
.
.
</form>
<?=$name?>
<br />
<a href="list_product.php">HOME</a>
<?
mysql_close($conn);
?>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<?php

$name = '<center><span style="font-size:9px;font-style:italic;"background-color:#119922;color:#993311;"></center><br />';
$id_product = $_REQUEST['id_product'];
require('conn_mysql.php');
$query = "use project"; 
$result = mysql_query($query) or die ("Query failed");

$query = "select * from product where id_product = '$id_product'";

$result = mysql_query($query) or die ("Query failed");
$row = mysql_fetch_array($result);
?>
<form action="up_product.php?id_product=<?=$id_product?>" method="post">
Edit product<br />
รหัสสินค้า : <input type="text" name="id_product"  value="<?=$row[id_product]?>" /><br />
ชื่อสินค้า : <input type="text" name="name_product" size="40" value="<?=$row[name_product]?>" /><br />
ราคา : <input type="text" name="price" value="<?=$row[price]?>" /><br />
จำนวนสินค้า : <input type="text" name="balance" value="<?=$row[balance]?>" /><br />
<br />
<input type = submit name = submit value = Submit>
<input type = reset name = reset value = Cancel>
</form>
<?=$name?>
<br />
<a href="list_product.php">HOME</a>
<?
mysql_close($conn);
?>