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";?>

回复讨论(解决方案)

$number1 这个变量没值嘛，$number1是从哪取的值呀。

$number1 这个变量没值嘛，$number1是从哪取的值呀。 谢谢，我突然发现我把变量名改了……这下好了