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";?>
$number1 这个变量没值嘛，$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
