当密码和用户名在数据库中匹配时，错误显示

Question

当在表单中写入任何内容并单击Enter按钮时，没有错误，但当我写入用户名和密码时，错误显示我尝试echo "$_POST['username']";//如果用户名不匹配，打印用户名

错误:

注意:未定义索引:用户名在..

注意:未定义索引:password in .

这是我的表单

<form action="2.php" method="post">

<table align="center">
    <tr>
    <td>Username</td>
    <td><input type="text" name="username" /></td>
    </tr>
    <tr>
    <td>Password</td>
    <td><input type="password" name="password" /></td>
    </tr>
    <tr>
    <td colspan="2" align="center"><input type="submit" value="Enter"  />
    </td>
    </tr>
</table>
</form>

这是我的第二页

<?php
$username = mysql_real_escape_string($_POST['username']); 
$password = mysql_real_escape_string($_POST['password']);
    $connection = mysql_connect('localhost', 'root', '');
    if (!$connection){
        die('Could not connect');
        exit; 
    }
    mysql_select_db('dbName') or die( "Unable to select database");
    $query = "SELECT * FROM admin WHERE username = '$username'";
    $result = mysql_query($query) or die(mysql_error());
    $num = mysql_num_rows($result); // number of rows
    if ($num > 0){
    $i = 0;
    while ($i < $num){

        $row = mysql_fetch_array($result);
        if ( ($password) == $row['password'] && ($username) == $row['username'] ){
            header('location:2.php');
            $_SESSION['sessionname'] = $username;
            $_SESSION['sessionpass'] = $password;
        }
        elseif ( ($password) != $row['password'] && ($username) == $row['username'] ) {
            echo "Wrong Password <a href='1.php' >Click Here</a>";
        }
        $i++;
        }
    }else {
        echo "Username  <strong><u>$_POST[username]</u></strong> invalid ! <a href='1.php' >Click Here</a> ";

        }
    ?>

Answer 1

没有理由按照您的方式分配用户名和密码变量。只需将post数据分配给变量，就像您通常使用会话一样。