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

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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

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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