PHP高手求助
怎麽改才能在表裹显增加成功，并往上30级加3元，并在会员表裹有account加成premium<?
include ('header.php');
//$user = limpiar($_POST["user"]);
$user = $row['user'];
$pemail = limpiar($_POST["pemail"]);
$email = limpiar($_POST["email"]);
$laip = getRealIP();
?>
<h3>
<img border="0" src="images/orders.gif" align="absmiddle" width="32" height="32">
<span style="font-weight: bold">高级会员</span>
</h3>
<?
if (!isset($_POST['passcode']))
{//若passcode尚未配置
?>
<div align="center">
<div id="form">
<fieldset><legend></legend>
<table width="302" border="0" cellpadding="3" cellspacing="1" bordercolor="#009900" bgcolor="#009900">
<tr>
<td bgcolor="#FFFFFF">
<div align="center" style="font-weight: bold">高级会员</div>
</td>
</tr>
<tr>
<td width="292" bgcolor="#FFFFFF">
<div align="left"><b>-3%</b> 30级团队提成</div>
</td>
</tr>
<tr>
<td bgcolor="#FFFFFF">
<div align="left"><b></b><b>-付款优先</b> - 通常得到支付不会超过2天</div>
</td>
</tr>
</table>
<?
$elus = $_COOKIE["usNick"];
require ('config.php');
$sql = "SELECT * FROM tb_users WHERE username='$elus'";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
mysql_close($con);
$dep1 = $row["username"];
$dep2 = $row["pemail"];
$dep3 = $row["email"];
$dep4 = $row["lastiplog"];
?>
<form method="post" action="upgrade.php"><input type="hidden"
name="user" value="<?php echo $dep1; ?>"> <input type="hidden" name="pemail"
value="<?php echo $dep2; ?>"> <input type="hidden" name="email"
value="<?php echo $dep3; ?>"> <input type="hidden" name="ip"
value="<?php echo $dep4; ?>"><br>
激活码 <input type="text" name="passcode" />
<p><input type="submit" value="升级" class="submit" tabindex="4" /></p>
</form>
</fieldset>
</div>
</div>
<?php
}
else
{//若passcode已经配置
if (empty($passcode))
{//若passcode为空
echo "<b>必须填写激活码!</b>";
?>
<tr>
<td width="150" align="left" style="height: 30px">
<p><label>激活码</label></p>
</td>
<td width="250" align="left" style="height: 30px"><input type="text"
size="25" maxlength="20" name="passcode" tooltipText="升级必需要激活 "
autocomplete="off" class="field" value="" tabindex="1"
style="width: 120px" /> <a
href="http://www.adpost2.com/bbs/read.php?tid=6"> <strong>索取</strong></a></td>
</tr>
<?php
}
else
{//若passcode非空
//查激活码
require ('config.php');
@mysql_select_db("dandan");
$passcode = $_POST["passcode"];
$passcode = limpiar($passcode);
$passcode = limitatexto($passcode, 15);
$checkpasscode = mysql_query("SELECT passcode FROM tb_code WHERE passcode='$passcode'");
$passcode_exist = @mysql_num_rows($checkpasscode);
mysql_close($con);
if ($passcode_exist == 0) {
echo "你输入的激活码不存在.";
include ('footer.php');
exit();
}
else
{
require ('config.php');
$sqle = "SELECT * FROM tb_upgrade WHERE username='$user'";
$resulte = mysql_query($sqle);
$rowe = mysql_fetch_array($resulte);
mysql_close($con);
if ($rowe["status"] == "upgraded") {
echo "Error: 请勿重复升级.";
include ('footer.php');
exit();
}
require ('config.php');
$query = "INSERT INTO tb_upgrade (username, pemail, email, ip) VALUES('$user','$pemail','$email','$laip')";
mysql_query($query) or die(mysql_error());
//计算奖金
//{
$ref = $referer; // 直接领导
$referers = array($referer); // 初始化领导数组
$count = 0; // 计数
$max = 30; // 最大层数
while (($rs = mysql_query("SELECT referer FROM tb_users WHERE username='$ref' LIMIT 1")) && $count < 30) { // 获取领导用户名
if (($row = mysql_fetch_array($rs)) && ! empty($row))
$referers[] = $ref = $row; // 添加到数组
$count ++;
}
// 更新所有领导的金额
$referers_str = count($referers) > 1 ? implode("','", $referers) : $referers; // 分解领导名为字符串 用 ',' 连接
$query = "UPDATE tb_users SET money=money+ 3 WHERE username IN ('$referers_str')";
mysql_query($query) or die(mysql_error());
}
$sqle = "SELECT * FROM tb_config WHERE item='upgrade' and howmany='1'";
$resulte = mysql_query($sqle);
$rowe = mysql_fetch_array($resulte);
mysql_close($con);
echo "升级成功!";
include ('paypal.php');
}
}
include ('footer.php');
?>
edited by iptton
asssssssssssss
$sql = "UPDATE tb_user SET account='premium' WHERE username='$user'";mysql_query( $sql ) or die( "upgrade fail." );
是这样改吗？
那个，其实，我还不是很明白LZ所表达的意思。。 MS是问SQL语法的吧？
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