登录表单脚本其他部分不工作 - Login form script else part not working

我为我的web应用程序写了一个简单的登录表单脚本。如果我的用户名和密码匹配，我可以登录到帐户。

我的问题是，如果输入的详细信息无效，它不会显示所需的消息。

脚本如下

<?php
include 'buildconfig.php';
$connection = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or
die("Failed to connect to MySQL:" . mysql_error());
$db=mysql_select_db(DB_NAME,$connection) or 
die("Failed to connect to MySQL: ".mysql_error());
function SignIn()
{
    // username and password sent from form
    $myusername=$_POST['username'];
    $mypassword=$_POST['password'];
    // 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);
    $sql="SELECT * FROM members WHERE uname='$myusername' and password='$mypassword'";
    $result = mysql_query($sql);
    if($result === FALSE) {
        die(mysql_error()); 
    }
    while($row = mysql_fetch_array($result))
    {
        if($row["uname"] == $myusername && $row["password"] == $mypassword)
        {
            $to_location = "Location:../pages/login.html";header($to_location);
        }
        //this else is not working
        else
        {
            echo"Sorry, your credentials are not valid, Please try again.";
        }
    }
}
if (isset($_POST['submit']))
{
    SignIn();
}
?>

我做错了什么吗?

或者有更好的方法来实现这一点?

当代码不匹配时，您将无法获得任何行，并使用mysqli_num_rows检查行是否存在。

if(mysql_num_rows($result)!==0)

使用

下面的代码

<?php
include 'buildconfig.php';
$connection = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or
die("Failed to connect to MySQL:" . mysql_error());
$db=mysql_select_db(DB_NAME,$connection) or 
die("Failed to connect to MySQL: ".mysql_error());
function SignIn()
{
// username and password sent from form
$myusername=$_POST['username'];
$mypassword=$_POST['password'];
// 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);
$sql="SELECT * FROM members WHERE uname='$myusername' and password='$mypassword'";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error()); 
}
if(mysql_num_rows($result)!==0){
while($row = mysql_fetch_array($result))
{
if($row["uname"] == $myusername && $row["password"] == $mypassword)
{$to_location = "Location:../pages/login.html";header($to_location);}
}
}
//this else is working
else
{echo"Sorry, your credentials are not valid, Please try again.";}
}

if (isset($_POST['submit']))
{SignIn();}
?>

希望这对你有帮助

TJ以上是正确的。错误回显在一个永远不会运行的while循环中。简单地把它拿出来，你甚至不需要if，因为你的头应该在到达else语句之前重定向if成功。

<?php
    include 'buildconfig.php';
    $connection = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or
    die("Failed to connect to MySQL:" . mysql_error());
    $db=mysql_select_db(DB_NAME,$connection) or 
    die("Failed to connect to MySQL: ".mysql_error());
    function SignIn(){
        // username and password sent from form
        $myusername=$_POST['username'];
        $mypassword=$_POST['password'];
        // 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);
        $sql="SELECT * FROM members WHERE uname='$myusername' and password='$mypassword'";
        $result = mysql_query($sql);
        if($result === FALSE) {
            die(mysql_error()); 
        }
        while($row = mysql_fetch_array($result)){
            if($row["uname"] == $myusername && $row["password"] == $mypassword){
                $to_location = "Location:../pages/login.html";
                header($to_location);
            }
        }
        // this will run after the while loop and in theory the header redirect so no need for else
        echo "Sorry, your credentials are not valid, Please try again.";
    }
    if (isset($_POST['submit'])){
        SignIn();
    }
?>