Question

我试图在用户点击提交时在同一页面上显示我的登录错误。当我在单独的文件上有PHP和HTML代码但我合并两个文件并使用

时，我的代码可以工作

<?php echo $_SERVER['PHP_SELF']; ?>

在action属性中，而不是给出文件位置，它只显示

die("Incorrect Username or Password entered");

错误。我完全不知道为什么会这样。

      <?php
ob_start();
include ("cn.php");
// Define $myusername and $mypassword
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$date = date("Y-m-d H:i:s");

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);

$sql="SELECT * FROM spineless.Users WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);
$user_info = mysql_fetch_assoc($result)
or die ("Incorrect Username or Password entered");
extract ($user_info);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{

// Register $myusername, $mypassword and redirect to file "joblist.php"
    session_register("myusername");
    session_register("mypassword");
    $_SESSION['myusername'] = $myusername; 
    $_SESSION['mypassword'] = $mypassword;
    $_SESSION['userid'] = $User_ID; 

    $user_record = "INSERT INTO Login_Record (User_ID, Username, Login_Time)
    VALUES
    ('$User_ID','$Username','$date')";
    $recordresult = mysql_query($user_record)
    or die ("unable to add record");

    header("location:../views/joblist.php");    

    //echo "yes";

}

else
{
    echo "Wrong Username or Password";
}
ob_end_flush();
?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Spineless Classics</title>
<link rel="stylesheet" type="text/css" href="css/stylesheet.css" />
</head>
<body id="loginPage">
    <div class="loginContainer">

        <div class="loginHolder">
            <div class="block">


                <div style="text-align:center; padding-bottom: 20px;"><a href="/" title=""><img src="img/spinelessclassics.png" ></a></div>
                <!--<div class="login-error">
                        Please enter your username and password</a> // HIDE AND DISPLAY
                    </div>-->
            <!-- /error_holder -->
                <form  name="login_form" id="login_form" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
                    <input type="myusername" name="myusername" placeholder="Username" class="login-input" mouseev="true" keyev="true" clickev="true" >
                    <input type="password" name="mypassword" placeholder="Password" class="login-input" mouseev="true" keyev="true" clickev="true">
                    <button type="submit" name="Submit" class="login-submit">Login</button>
                </form>


            </div>
        </div>



    </div>








</body>
</html>

Answer 1

用以下代码包装整个PHP代码：

if(isset($_POST['Submit']){
 //all your PHP code
}

这样做的原因是，当您打开页面时，它会检查尚不存在的POST值（，因为您尚未提交表单）。

isset() 函数检查是否设置了提交的值，然后才应开始处理POST值。

注意：

Please, don't use mysql_* functions in new code。它们不再被维护and are officially deprecated。请参阅red box？转而了解prepared statements，并使用PDO或MySQLi - this article将帮助您确定哪个。如果您选择PDO here is a good tutorial。

在同一页面上显示登录错误

1 个答案:

注意：