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";?>
------解决方案--------------------
$number1 这个变量没值嘛,$number1是从哪取的值呀。