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Code (PHP)
<?php
include 'conn.php';
$sql_edit = "UPDATE raw_material SET rm_name= '".$_POST['name_rm']."',price_raw ='".$_POST['price_rm']."' WHERE rm_id = '".$_GET['ID']."'";
$sql_edit = "UPDATE have SET";
$sql_edit .= "nutrients = '".$_POST["EN"]."' ";
$sql_edit .= ",nutrients = '".$_POST["CP"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Cal"]."' ";
$sql_edit .= ",nutrients = '".$_POST["P"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Arg"]."' ";
$sql_edit .= ",nutrients = '".$_POST["His"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Iso"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Leu"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Lys"]."' ";
$sql_edit .= ",nutrients = '".$_POST["MC"]."' ";
$sql_edit .= ",nutrients = '".$_POST["PT"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Thr"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Trp"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Val"]."' ";
$sql_edit .= ",nutrients = '".$_POST["GC"]."' ";
$sql_edit .= ",nutrients = '".$_POST["Lin"]."' ";
$sql_edit .= "WHERE rm_id = '".$_GET['ID']."'";
$objQuery = mysql_query($sql_edit) or die (mysql_error());
?>
<script>
alert('แก้ไขข้อมูลวัตถุดิบเรียบร้อยแล้วค่ะ');
window.history.back();
</script>
<?php
mysql_close();
?>
ขึ้น error แบบนี้อ่าค่ะ

Tag : PHP, MySQL
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Date :
2014-08-16 18:32:14 |
By :
จุ๋มจิ๋ม |
View :
669 |
Reply :
4 |
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