修改信息成功后内容未变
系统有4处错误提示，分别在第15、17、25、26行。Undefined variable number1（PS：我发现我一个月都在问这个问题……惭愧）

<?php
   include "conn.php";
   include "admin_header.php";
   extract($_REQUEST);
  
   $query="update $jiaoshi_table set     subject='$newsubject',teacher='$newteacher',zhicheng='$newzhicheng',specialized='$newspecialized',code='$newcode',intrduction='newintroduction' where id='$id'";
   mysql_query("set names 'GB2312'");
   $result=mysql_query($query);
   
   
   $query="select number as sn,surplus as ssn from $jiaoshi_table where id='$id'";
   mysql_query("set names 'GB2312'");
   $result=mysql_query($query);
   $row=mysql_fetch_array($result);
if($number1<$row['sn'])
{
if($number1<($row['sn']-$row['ssn']))
{
echo"<p align=\"center\"><font color=\"#FF0000\"><b><big>人数不能少于现已选题人数，人数列修改失败！</big></b></font>";
echo"<meta http-equiv=\"refresh\" content=\"2;url=alter_keti.php\">";
exit;
}
else
{
    $query2=mysql_query("update $jiaoshi_table set number='$number1' where id='$id'");
$query3=mysql_query("update $jiaoshi_table set surplus=surplus-($row[sn]-$number1) where id='$id'");
mysql_query("set names 'GB2312'");
   $result1=mysql_query($query3);
}
}
else
{
$query4=mysql_query("update $jiaoshi_table set number='$number1' where id='$id'");
$query5=mysql_query("update $jiaoshi_table set surplus=surplus+($number1-$row[sn]) where id='$id'");
mysql_query("set names 'gb2312'");
   $result2=mysql_query($query5);
   }
if($result==true)
{
   echo"<p align=\"center\"><font color=\"#FF0000\"><b><big>修改课题成功！</big></b></font>";
   echo "<meta http-equiv=\"refresh\" content=\"1;url=alter_keti.php\">";
   exit;
 }
 else
 {
   echo"<p align=\"center\"><font color=\"#FF0000\"><b><big>修改出错，请返回重新修改</big></b></font></p>";
   echo "<meta http-equiv=\"refresh\" content=\"1;url=alter_keti.php\">"; 
   exit;
 }

?>
<?php include "foot.php";?>

[解决办法]
$number1 这个变量没值嘛，$number1是从哪取的值呀。