请教一下，这个php程序究竟有什么问题啊？看了很多遍都没有发现问题的所在

Question

<?php
session_start();
if(! isset($_SESSION['name'])){
   if(isset($_POST['usename'])){
      $username=$_POST['username'];
  $pswd = $_POST['pswd'];
  mysql_connect("localhost","root","");
  mysql_select_db("student_db");
  $query="SELECT name,username FROM userauth";
  $query.="WHERE username='$username' AND passwd=PASSWORD('$pswd')";
  mysql_query("SET NAMES 'gbk'");
  $result = mysql_query($query);
      if(mysql_num_rows($result)==1){
     $_SESSION['name']=mysql_result($result,0,"name");
 $_SESSION['username']=mysql_result($result,0,"username");
 echo $_SESSION['name'].",你成功登录了。<br>";
     }   else{
     include "login.html";
 exit;
     }
  else{
    include "login.html";
exit;
  }
}else{
 $name = $_SESSION['name'];
 echo "欢迎回来,$name!<br>";
}
?>
<a href="logout.php">退出</a>

Answer 1

! isset($_SESSION['name']这个空格是？

Answer 2

最明显的错误是第二个if语句没有结束的“}“。<?php
session_start();
if(! isset($_SESSION['name'])){
if(isset($_POST['usename'])){
$username=$_POST['username'];
$pswd = $_POST['pswd'];
mysql_connect("localhost","root","");
mysql_select_db("student_db");
$query="SELECT name,username FROM userauth";
$query.="WHERE username='$username' AND passwd=PASSWORD('$pswd')";
mysql_query("SET NAMES 'gbk'");
$result = mysql_query($query);
if(mysql_num_rows($result)==1){
$_SESSION['name']=mysql_result($result,0,"name");
$_SESSION['username']=mysql_result($result,0,"username");
echo $_SESSION['name'].",你成功登录了。<br>";
}
else{
include "login.html";
exit();
}
}
else{
include "login.html";
exit();
}
}
else{
 $name = $_SESSION['name'];
 echo "欢迎回来,$name!<br>";
}
?>
<a href="logout.php">退出</a>

Answer 3

在第二个else前面加个"}",这样结构才完整，楼主先测试下！至于有其他错误再说