Question

所以我的登录工作，但试图回应数据库中的用户名不会工作错误说＆＃34;解析错误：语法错误，意外$ end在/home/..../public_html/app/views/login/logout.php第31行和第34行; 我的代码：

  <?php  
    session_start();

    if (isset($_POST['username']) and isset($_POST['password'])){
    //3.1.1 Assigning posted values to variables.
    $username = $_POST['username'];
    $password = $_POST['password'];

    $query = "SELECT * FROM `loginuser` WHERE username='$username' and password='$password'";

    $result = mysql_query($query) or die(mysql_error());
    $count = mysql_num_rows($result);
    //3.1.2 If the posted values are equal to the database values, then session will be created for the user.
    if ($count == 1){
    $_SESSION['username'] = $username;
    }else{
    //3.1.3 If the login credentials doesn't match, he will be shown with an error message.
    echo "Invalid Login Credentials.";
    }
    }

    if (isset($_SESSION['username'])){
    $username = $_SESSION['username'];
    echo "Hello " . $username . "";
    echo "This is the Members Area";
    echo "<a href='logout'>Logout</a>";
    }else{


    ?>

Answer 1

您最后需要为}语句添加else。或删除它，因为它没有被使用。

}else{
}
?>

Answer 2

<?php  
session_start();

if (isset($_POST['username']) and isset($_POST['password'])){
//3.1.1 Assigning posted values to variables.
$username = $_POST['username'];
$password = $_POST['password'];

$query = "SELECT * FROM `loginuser` WHERE username='$username' and password='$password'";

$result = mysql_query($query) or die(mysql_error());
$count = mysql_num_rows($result);
//3.1.2 If the posted values are equal to the database values, then session will be created for the user.
if ($count == 1){
$_SESSION['username'] = $username;
}else{
//3.1.3 If the login credentials doesn't match, he will be shown with an error message.
echo "Invalid Login Credentials.";
}
}

if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "Hello " . $username . "";
echo "This is the Members Area";
echo "<a href='logout'>Logout</a>";
}else{
    // Do nothing
}
?>

或者完全忽略最后一个。

登录回显不起作用

2 个答案: