会PHP的帮我看看阿~这个那错了阿~~~~~
[code]<?if (isset($_GET["id"]))
{
$id=$_GET["id"];
$zcid=$_GET["zcid"];
$zzjj=$_POST["zcjj"];
$tablae = mysql_query("SELECT * FROM tb_zhu where id='$zcid'");
while ($registroe = mysql_fetch_array($tablae)) {
$lolzeaq=$registroe["zctjsl"];
$moneyero= $lolzeaq - 1;
$query = "UPDATE tb_zhuce SET zctjsl='$moneyero' where id='$zcid'";
mysql_query($query) or die(mysql_error());
$query = "UPDATE tb_zhucetj SET zcshzt='0',zcjjyy='$zcjjyy' where id='$id'";
mysql_query($query) or die(mysql_error());
echo "<font color=\"green\"><b>更新成功!</b></font><br><br>";
}
if (isset($_GET["id"]))
{
$id=$_GET["id"];
$option=$_GET["option"];
if ($option=="jujue"){
?>
<form method="post" action="index.php?op=38">
你要写入的备注: <input type="text" name="zcjjyy" value=""><br>
<input type="submit" value="保存" class="button">
</form>
<br><br>
<?
}
}
?>
<br><br>
<table>
<tr>
<th>序号</th>
<th>相关ID</th>
<th>用户名</th>
<th>邮箱地址</th>
<th>注册时间</th>
<th>IP</th>
<th></th>
<th></th>
</tr>
<?
$tabla = mysql_query("SELECT * FROM tb_zhucetj where zcshzt='' ORDER BY id ASC");
while ($registro = mysql_fetch_array($tabla)) {
echo "
<tr>
<td>". $registro["id"] ."</td>
<td>". $registro["zcid"] ."</td>
<td>". $registro["username"] ."</td>
<td>". $registro["email"] ."</td>
<td>". $registro["joindate"] ."</td>
<td>". $registro["ip"] ."</td>";
?>
<td>
<form method="post" action="index.php?op=38&id=<?= $registro["id"] ?>&zcid=<?= $registro["zcid"] ?>&option=jujue" >
<input type="submit" value="提交" class="button">
</form>
</td>
</tr>
<?
} // fin del bucle de ordenes
?>
</table>[/code]我每次按提交后直接执行了[code]$id=$_GET["id"];
$zcid=$_GET["zcid"];
$zzjj=$_POST["zcjj"];
$tablae = mysql_query("SELECT * FROM tb_zhu where id='$zcid'");
while ($registroe = mysql_fetch_array($tablae)) {
$lolzeaq=$registroe["zctjsl"];
$moneyero= $lolzeaq - 1;
$query = "UPDATE tb_zhuce SET zctjsl='$moneyero' where id='$zcid'";
mysql_query($query) or die(mysql_error());
$query = "UPDATE tb_zhucetj SET zcshzt='0',zcjjyy='$zcjjyy' where id='$id'";
mysql_query($query) or die(mysql_error());
echo "<font color=\"green\"><b>更新成功!</b></font><br><br>";[/code]这段,
使得我在要写入备注的地方直接跳过去了!
我想的是按提交后,显示出要那个要我输入备注的地方,然后再次提交,并把这些内容更新数据库!
那个高手帮我下阿! :mad: :mad: :mad: :mad: :mad: :mad: 感觉没有问题，呵呵
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