通过JavaScript检查php会话，如果不存在则不会导致任何错误

Question

var secret_photo
var secret_name

if (!<?php echo isset($_SESSION['fb_item'])?'true':'false'; ?>) {
if($.cookie("domain[user]")){
secret_photo = 'noavatar.png';
secret_name = 'admin';
}
} else {
secret_photo = <?php echo json_encode($_SESSION['fb_item']['url']); ?>;
secret_name = <?php echo json_encode($_SESSION['fb_item']['name']); ?>;
}

I need to set php session to html using JavaScript .There is 2 logins , one is the facebook and the other is my own login system, facebook uses session and my own login sets cookie.

When I login by facebook , things are good , but if I login using my own login system , that echo check is going to cause problem , so how am I going to fix it ? Check if the session is available or not even if it doesn't exist it wont cause any problem.

Answer 1

You are probably encounter an error at the bottom of your code. You can't access $_SESSION['fb_item'] if it doesn't exist, it either shows a Notice or generate wrong js code secret_photo = ; . Just move condition to php side:

var secret_photo
var secret_name

<?php if(!isset($_SESSION['fb_item'])):?>
    if($.cookie("domain[user]")){
        secret_photo = 'noavatar.png';
        secret_name = 'admin';
    }
<?php else:?>
    secret_photo = <?php echo json_encode($_SESSION['fb_item']['url']); ?>;
    secret_name = <?php echo json_encode($_SESSION['fb_item']['name']); ?>;
<?php endif;?>

Answer 2

<script type="text/javascript">
var secret_photo
var secret_name
<?php if(!isset($_SESSION['fb_item'])){?>
    if($.cookie('user')){
        secret_photo = 'noavatar.png';
        secret_name = 'admin';
    }
<?php }else{?>
    secret_photo = <?php echo json_encode($_SESSION['fb_item']['url']); ?>;
    secret_name = <?php echo json_encode($_SESSION['fb_item']['name']); ?>;
<?php }?>
</script>