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在我的网站中,当我们从索引页面登录时,登录功能可以正常工作。但是当我们从我网站的索引页面以外的其他页面登录时,它没有登录。

http://domain.com/mysite/index.php

从该页面登录时,该功能正常工作,登录后显示如下页面

登录成功后

如果我们从我网站中的其他页面登录,它将无法正常工作,并且页面显示为 http://mysite.com/samplefolder/detailpage.php?pollid=2&chid=2

登录后也显示登录

登录.php

<?php
include("jalert.php");
if (array_key_exists("login", $_GET)) {

$oauth_provider = $_GET['oauth_provider']; if ($oauth_provider == 'twitter') { header("Location: login-twitter.php"); } } ?>

<script type="text/javascript">

$(document).ready(function() {
$("#new_user_session_ajax").submit(function(e){
    e.preventDefault();
       var email = $('#userid').val();
       var password = $('#signin_password').val();
       var remember = $('#user_session_remember_me').val();

        $.ajax({
        type: "POST",
        url: "ajaxlogin.php",
        data: "userid="+email+"&password1="+password+"&rememberme="+remember+""
        }).done(function( msg )  {
    if(msg)
    {
        $('#error').html(msg);
    }
    else
    {
        window.location.href="index.php";
    }

});
});
});

</script>


<form accept-charset="UTF-8" action="" name="login" onsubmit="return checkreg();" class="no_scroll_animation validate no_bump ajax" id="new_user_session_ajax" method="post">
<input name="utf8" type="hidden" value="">
<input name="authenticity_token" type="hidden" value="yRcKhChgROcxunh7A7zRQVroShpYbO6e6PaAO4JHUtg=">
</div>

<input id="after_url" name="redirect_after_login" type="hidden" value="/">


<li class="remember_me clear">
<input name="user_session[remember_me]" type="hidden" value="0">
<input checked="checked" id="external_remember_me_popup" name="remember_me" type="checkbox" value="1">
<label for="external_remember_me_popup">Remember me</label>


<h5>Sign in with email</h5>


<li><input autocapitalize="off" class="validate validate_required validate_email placeholder" id="userid" name="userid" value="Email Address"
           onfocus="if(this.value == 'Email Address') {this.value=''}" onblur="if(this.value == ''){this.value ='Email Address'}"  title="email" type="email" value=""></li>

<li><input autocapitalize="off" class="validate validate_required placeholder" id="signin_password" name="password1" value="Password"
           onfocus="if(this.value == 'Password') {this.value='';this.type='password'}" onblur="if(this.value == ''){this.value ='Password';this.type ='text'}" type="text" validate:message="Please enter your password" value=""></li><li class="last">


    <input class="button" name="login" type="submit" value="Login"  >
<div class="keepmeloggedin"><input name="rememberme" type="hidden" value="remember me"><input checked="checked" id="user_session_remember_me" value="remember me" name="rememberme" type="checkbox" value="1">&nbsp; Remember me</div><div class="forgotPassword">
<a class="forget_pass" href="forgot_password.php" id="forgotPasswordLink" rel="nofollow">Forgot password?</a></div></li></ul></div></form>

ajaxlogin.php

<?php 
include("classes.php");
$jeob = new EF_Sql();

    $email= $_POST['userid'];

    $password= $_POST['password1'];

    $papas=base64_encode($password);

    $check = $_POST['rememberme'];



        $tablename="user_record";

    $select_qry = $jeob->SqlQuery("SELECT * FROM ".$jeob->dbprefix.$tablename." WHERE email ='$email' AND password ='$papas' AND active_link='1' ");

$kiss="SELECT * FROM ".$jeob->dbprefix.$tablename." WHERE email ='$email' AND password ='$papas'";

        if($jeob->SqlRows($select_qry) == "0"){                 

         echo "Invalid Username and Password";


        } else {


        if($check) {

                setcookie("username", $email, time() + 3600);       // Sets the cookie username

        }

        $getuser = $jeob->SqlFetch($select_qry);

        $_SESSION['userid'] = $getuser['user_id'];  

        $_SESSION['oauth_provider'] = "normal";

        $_SESSION['email'] = $getuser['email'];     

        }
?>
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1 回答 1

0

确保您session_start();在每个文件的顶部调用。您可能在某些页面上遗漏了该内容,因此会话不会继续,因此看起来好像用户已注销。

于 2013-01-02T05:55:18.333 回答