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我有一个验证登录脚本的 AJAX 脚本，当用户输入不正确的数据时，会出现错误消息，但是当用户输入正确的数据时，新页面会加载到错误消息空间中。我知道问题出在哪里，我只是不知道如何解决它。

下面是登录脚本和检查登录脚本

登录.php

<table class="loginTable" align="center">
    <tr>
        <td>
            Email:
        </td>
        <td>
            <input type="text" name="email" id="email"/>
        </td>
    </tr>
    <tr>
        <td>
            Password:
        </td>
        <td>
            <input type="password" name="password" id="password"/>
        </td>
    </tr>
    <tr>
        <td align="center" colspan="2">
            <span id="ErrorMessage"></span>
        </td>
    </tr>
    <tr>
        <td colspan="2" align="center">
            <!--Keep me logged in? <input type="checkbox" name="keeploggedin" /><br /><br />-->
            <input  id="loginButton" type="button" name="login" value="Login" onclick="processLogin()"/>
        </td>
    </tr>
</table>


<script>

    function processLogin()
{
var xmlhttp;

var email = document.getElementById("email").value;
var password = document.getElementById("password").value;

if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function() 
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("ErrorMessage").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("POST","checklogin.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("email=" + email + "&password=" + password);
}

</script>

检查登录.php

<?php

session_start();

//Login Script



//Variables
$email = $_REQUEST['email'];
$userPassword = $_REQUEST['password'];
//$keeploggedin = $_REQUEST['keeploggedin'];



require_once("dbdetails.php");

// Create Mysqli object
$db = new mysqli($hostname, $username, $password, $database);

// Create statement object
$stmt = $db->stmt_init();

// Create a prepared statement
if($result = $stmt->prepare("SELECT u.UserFirstName, u.UserSurname, u.UserID FROM user u WHERE u.UserEmail = ? AND u.UserPassword = ?")) {

    // Bind your variables to replace the ?s
    $stmt->bind_param('ss', $email, $userPassword) or die(errorCodes($stmt->errno));

    //$row -> fetch_array(MYSQLI_ASSOC);

    // Execute query
    $stmt->execute() or die(errorCodes($stmt->errno));

    $stmt->bind_result($fname, $sname, $userID);

    $stmt->store_result();

    $count = $stmt->num_rows;

    if($count == 1){
        while($row = $stmt->fetch()) {

            // Register $email, $password, $firstname, $surname, create logged_in session and redirect to file "projects.php"
            $_SESSION['email'] = $email;
            $_SESSION['password'] = $userPassword;
            $_SESSION['firstname'] = $fname;
            $_SESSION['surname'] = $sname;
            $_SESSION['userID'] = $userID;
            $_SESSION['LoggedIn'] = 'true';
            $_SESSION['loginCount'] = 0;
            header("location:projects.php");
        }
    }
    else {
        //$_SESSION['loginCount'] += 1;
        //header("location:loginPage.php");
        //echo("<script type='text/javascript'>\n");
        //echo("changeError();\n");
        //echo("</script>");    

        echo("Username and Password mismatch, please try again");
    }

    // Close statement object
    $stmt->close();
}


    function errorCodes($aErrorCode)    {
        echo("Error " . $aErrorCode . " has occured");
    }


?>
4

3 回答 3

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登录成功时返回一些 json 数据，说明登录成功（来自 php），然后修改您的此代码

xmlhttp.onreadystatechange=function() 
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("ErrorMessage").innerHTML=xmlhttp.responseText;
    }
  }

success如果您从 php.ini 获得响应，则将用户重定向到下一页。

于 2013-03-19T12:47:29.000 回答
0

您可以测试xmlhttp.responseText：

  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
       if(xmlhttp.responseText != "")    //If not, it completes your error box
          document.getElementById("ErrorMessage").innerHTML=xmlhttp.responseText;
       //Else, let the redirection trigger

    }
于 2013-03-19T12:50:25.003 回答
0

使用简单的ajax，希望对您有所帮助:)

<script>
function processlogin()
{
$.ajax(
{
url:'checklogin.php',
type:'POST',
data:{'email':email,'password':password},
success:function(xyz)
{
if(xyz==1)
{
$("#errormessage").load("ajax/newcontent.php");
}
else{
//do something as you need
}
}
});
}
</script>

in the checklogin.php page, if the IF condition success then 'echo 1' or 'echo 0' in else part.so that you can validate using these values in success block of the ajax script.
于 2013-03-19T13:06:21.230 回答