1

当在表单中写入任何内容并单击 Enter 按钮时，没有错误，但是当我输入用户名和密码时，错误出现了，我尝试 echo "$_POST['username']"; // 如果用户名不匹配，则打印用户名？

错误：

注意：未定义的索引：用户名在..

注意：未定义索引：密码在..

这是我的表格

<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> ";


        }
    ?>
4

1 回答 1

1

没有理由像您那样分配用户名和密码变量。只需像通常使用会话一样将发布数据分配给变量。

于 2013-05-11T17:29:43.440 回答