Forums / Developer / How to know if user is logged in from inside template code

How to know if user is logged in from inside template code

Author Message

Michael Scofield

Tuesday 15 July 2008 9:08:39 pm

Hi all,

Sorry for the simple question, but I've googled a lot and couldn't find how to do it.

Inside a template, what is the best and faster way to know if a user is logged in at the website?

I want to limit some functionallity of my website to logged in users.

In a website made without eZ Publish we probably could check it using $_SESSION variables. What about with eZ Publish?

Btw, I found a fetch function that appears to do that:

{if fetch( 'content', 'access', hash( ...) ) }

{/if}

Should I use that or there is a better solution?

Thank you
Michael Scofield

Bård Farstad

Tuesday 15 July 2008 11:29:19 pm

You can try this template code:

{if $current_user.is_logged_in}
LOGGED IN
{else}
NOT LOGGED IN
{/if}

Also remember to not have cache blocks around this code, or if you do remember to use the user_id part of the cache key.

-bård

Documentation: http://ez.no/doc

André R.

Wednesday 16 July 2008 1:29:41 am

Or if you only need to know if the user is logged in (no need for user id or user name), then use $current_user.is_logged_in as cache key.

Also note that inside node/system templates you'll need to fetch the user first, so use something like this first*:

{if is_unset( $current_user )}
    {def $current_user = fetch('user', 'current_user') }
{/if}

But if you do want to check if a user has access to a certain module / function, then fetch('content', 'access') is a cleaner solution.

* node templates use content view cache, witch is unique pr user rights, but not user id. So you have to disable view cache (search for cache_ttl) if you need to use name / id in a node template.

eZ Online Editor 5: http://projects.ez.no/ezoe || eZJSCore (Ajax): http://projects.ez.no/ezjscore || eZ Publish EE http://ez.no/eZPublish/eZ-Publish-Enterprise-Subscription
@: http://twitter.com/andrerom