Question

我目前正在制作一个登录表单，虽然我知道当我尝试登录时没有任何错误，但是，它没有显示任何成功的消息或任何其他因此我不知道我是否成功登录是否。

<?php
        session_start();
        require_once('sqlconnect.inc.php');

        if (isset($_POST["Login"]))
        {
        $conn = @mysqli_connect($host, $user, $pswd);
                    if (!$conn) {
                    echo "<p>Database connection failure</p>"; 
                    } else {

        $selectDatabase = @mysqli_select_db($conn,$dbnm)
                    or die("<p>The database is not available.</p>");
                    }

        $email = $_POST['email'];
        $passw = $_POST['password'];

        $query = "SELECT member_email FROM team WHERE member_email = '$email' AND password = '$passw'";

        $queryResult = @mysqli_query($conn,$query)
                        or die ("<p>Unable to execute query.</p>". "<p>Error code" . mysqli_errno($conn) .":" . mysqli_error($conn))."</p>";


        if(mysqli_num_rows($queryResult) == 0) //user is not found
        {
            header('Location: login.php');
        }else{

        if(mysqli_num_rows($queryResult) == 1)  
        {
            echo ("<p>User is found, Successful login!</p>");   
            echo('<p><a href="memberadd.php">member add</a> </p>');
            echo('<p><a href="logout.php">Log out</a> </p>');

            $query2 = "SELECT member_name FROM team WHERE member_email = '$email' AND password= '$passw'";

            $queryResult2 = @mysqli_query($conn, $query2)
                        or die ("<p>Unable to execute query.</p>". "<p>Error code" . mysqli_errno($conn) .":" . mysqli_error($conn))."</p>";


            $array = mysqli_fetch_row($queryResult2);       
            $_SESSION['membername'] = $array[0];
        }
        else
        {
            echo"<p>Email and password do not match</p>";
            echo'<p><a href="index.php">Home page</a> </p>';
        }
        }
        }


?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/chtml-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en" >
<head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8" />
    <meta name="description" content="Web Programming :: Assignment 2" />
    <meta name="Keywords" content="Web, programming" />
    <title>Register Page</title>
</head>

<body>

<form id='login' action='login.php' method='POST'>
        <fieldset >
            <legend><h1>My Team System Log in Page</h1></legend>
            <?php $email = isset($_POST['email']) ? filter_var($_POST['email'], FILTER_SANITIZE_STRING) : ''; ?>
            <label for='email' >Email:</label>
                <input type='text' name='email' id='email' maxlength="50"  value="<?php echo $email; ?>" />
            </div>
                    <br />
            <div class="elements">
            <label for='password' >Password:</label>
                <input type='password' name='password' id='password' maxlength="50" />
            </div>
                    <br />
            <div class="submit">
                <input type='submit' name='login' value='Login' />
                <input type='reset' name='Submit' value='Clear' />
                    <br />
            <div class="elements">
            <a href="index.php">Home</a> 
        </fieldset>
</form>

</body>
</html>

Answer 1

要么通过使用@或代码块来屏蔽mysql调用中的错误，要么调用头函数。如果你删除@和/或用简单的

替换标题调用会发生什么

echo "Debug: Test";

登录表单不会显示任何成功的消息

1 个答案: