Question

我对PHP有点生疏。我想要做的是保存2个变量以在我的项目中使用。我想保存登录页面中的电子邮件和密码，以便稍后在Parse.com上解析我的数据库。

使用参考链接：[http://php.net/manual/en/book.session.php]

已解决，谢谢@Fred -ii-的帮助和解释！

登录文件：

try{

$query = new ParseQuery("Clients");
$query->equalTo("email", "$email");
$query->equalTo("password","$password");
$results = $query->find();

} catch(Exception $e){
echo $e->getMessage();

}
if(count($results)==1){
session_start();
$_SESSION['email'] = $email;
$_SESSION['password'] = $password;
 header("Location: http://localhost/ClientPortal/mainMenu.php");
}

mainMenu文件：

   session_start();
   $email = $_SESSION['email'];
   $password = $_SESSION['password']; //* fixed spelling error (used to be 'passowrd')
   session_commit();

try{

   $query = new ParseQuery("Clients");
   $query->equalTo("email", "$email");
   $query->equalTo("password","$password");
   $results = $query->find();

   } catch(Exception $e){
  echo $e->getMessage();

   }

if(count($results)==1){
echo "<p align=center>1 correct row</p>";
}
else{
 echo "<p align=center>NO correct rows</p>";
}

Answer 1

谢谢@Fred -ii-帮助我解决这个问题!!!

登录文件：

try{

$query = new ParseQuery("Clients");
$query->equalTo("email", "$email");
$query->equalTo("password","$password");
$results = $query->find();

} catch(Exception $e){
echo $e->getMessage();

}
if(count($results)==1){
session_start();
$_SESSION['email'] = $email;
$_SESSION['password'] = $password;
 header("Location: http://localhost/ClientPortal/mainMenu.php");
}

mainMenu文件：

session_start();
$email = $_SESSION['email'];
$password = $_SESSION['password']; //* fixed spelling error (used to be 'passowrd')
session_commit();

try{

  $query = new ParseQuery("Clients");
  $query->equalTo("email", "$email");
 $query->equalTo("password","$password");
 $results = $query->find();

 } catch(Exception $e){
 echo $e->getMessage();

 }

if(count($results)==1){
echo "<p align=center>1 correct row</p>";
 }
 else{
 echo "<p align=center>NO correct rows</p>";
}

PHP使用会话在项目文件中保存两个变量

1 个答案: