2016-11-13 18:43
浏览 71


I have a simple login form with checbox remember me. What i want to achieve is when user checked checbox than create cookies else start session but it still start session even I checked checkbox. Can anyone help ?

This is my html:

<form action="PHP/login.php" method="post" class="loginForm">

                <div class="form_vkope">
                    <input type="text" name="prihlNick" placeholder="Nick" class="prihlNick" />
                    <img src="Obrazky/ios7-person.png" alt="Ikonka postavy">                    

                <div class="form_vkope">
                    <input type="password" name="prihlHeslo" placeholder="Heslo" class="prihlHeslo" />
                    <img src="Obrazky/locked.png" alt="Ikonka zámok">                   

                <div class="obal_submitov">
                    <input type="checkbox" name="zapametat" class="zamapetat" /><label for="zapametat"><span></span>Remember me</label>
                    <input type="submit" name="prihlasit" value="Login" class="prihlasit" />                    

part of my PHP:

$zapametat = $_POST['zapametat']; //checkbox
 if ($dbNick == $Snick AND $dbHeslo == $Sheslo) {
        if ($ban != 1) {
            if (isset($zapametat)) {
                setcookie('id',$dbId,time()+86400, '/');
                setcookie('nick',$dbNick,time()+86400, '/');
                echo "cookie";
                $_SESSION['id'] = $dbId;
                $_SESSION['nick'] = $dbNick;
                echo "session";
            echo "Tvoj účet bol zablokovaný";
        echo "Heslo alebo meno sa nezhoduje";

And my jQuery:

$('.prihlasit').click(function() {
    var prihlNick = $('.prihlNick').val();
    var prihlHeslo = $('.prihlHeslo').val();
    var prihlasit = $('.loginForm .prihlasit');
    var checbox = $('.zapametat'); //checkbox
    var data = 'prihlNick='+prihlNick+'&prihlHeslo='+prihlHeslo+'&zapametat='+checbox+'&prihlasit='+prihlasit;
    if (prihlNick == '' || prihlHeslo == '') {
        $('.loginForm :input').addClass('inputError');
        $('.loginForm :input').removeClass('inputError');
            url: 'PHP/login.php',
            type: 'POST',
            data: data ,
        .done(function(data) {

1条回答 默认 最新

相关推荐 更多相似问题