这是以下代码。我试图创建一个工具与网站，这样我就可以输入项目到SQL数据库vs.使用GMTool的游戏，这是一个单独的工具。我可以手动输入信息，它100%工作。但是，当我尝试使用以下代码时，它说它是成功的，但是，数据库中没有发布任何信息。有什么想法吗?
<?php
require_once('config.php');
require_once('config/auth.php');
?>
<?php include("head.php"); ?>
<?php include("menu.php"); ?>
<?php if($logeado == 1){?>
<td id="column2">
<div id="title"><h2>Give Item</h2></div><div id="address">HOME > Account and Item</div>
<div id="content">
<hr>
Note - The world is 0 for ZeNFinix<br>
<form method="POST" action="#">
<input Value="World" id="world" name="world" vaulue="">
<input Value="Account" id="account" name="account" vaulue="">
<input Value="Item Number" id="index" name="index" vaulue="">
<input Value="Item Count" id="number" name="number" vaulue="">
<input type="Submit" Value="Continue" OnClick="if (! confirm('Are you sure you want to continue?')) return false;">
<input type="hidden" name="action" value="submitted">
</form>
<br><br>
<?php
if ($_POST['action'] == 'submitted')
{
global $link;
mssql_select_db("UserLogin",$link);
$acct = $_POST['acct'];
$world = $_POST['world'];
$userid = $_POST['account'];
$item = $_POST['index'];
$itemno = $_POST['number'];
$DateTime = 'GETDATE()';
$querys = mssql_query("SELECT * FROM dbo.Account WHERE userId = '$userid' ");
$QRow = mssql_fetch_array($querys);
if($QRow[0] == ''){
echo "Account does NOT Exist!";
}
else
{
{
}
global $link;
mssql_select_db("Tantra",$link);
mssql_query ("INSERT INTO dbo.TantraItem(World, Account, ItemIndex, ItemCount, DateTime) VALUES ('$world', '$userid', '$item', '$itemno', '$DateTime'");
echo "Successfully Added Item(s)!";
}
}
?>
</div>
</div>
</td>
</tr>
</table>
<?php } ?>
<?php include("footer.php"); ?>
"这是SQL数据库。dbo.TantraItem。"设计如下:ID int不允许为空World int不允许为空帐户varchar(50)不允许为空ItemIndex int不允许为空ItemCount smallint不允许为空DateTime允许为空
修改Mssql查询
mssql_query ("INSERT INTO dbo.TantraItem(World, Account, ItemIndex, ItemCount, DateTime) VALUES ('$world', '$userid', '$item', '$itemno', '$DateTime'");
mssql_query ("INSERT INTO dbo.TantraItem(World, Account, ItemIndex, ItemCount, DateTime) VALUES ('$world', '$userid', '$item', '$itemno', '$DateTime')");