how to check if this user is logged in?

Next topic

Author	Message
zaxofeel .v	Sunday 10 July 2005 3:02:22 am hey ppl, how are you all? :) as usual, i need ur help, i want to check if the current user is logged in or not, so, i found this piece of code in one of the ez templates: {section show=eq($current_user.contentobject_id,$anonymous_user_id)} <a class="menuitem" href={"/user/login/"\|ezurl}>{"Login"\|i18n("design/standard/layout")}</a> {section-else} <a class="menuitem" href={"/user/logout/"\|ezurl}>{"Logout"\|i18n("design/standard/layout")}</a> ({content_view_gui view=text_linked content_object=$current_user.contentobject}) {/section} when i used this code, i found that it always displays "login", and it never displayes logout even if i am already signed in. i really hope if someone can tell me what's wrong with that or if there is another way to implement this, i'll be very thankful. thanks all
Pål J Didriksen	Sunday 10 July 2005 5:48:07 am Hi! I think you need to fetch the current user first, for example: {default current_user=fetch('user','current_user')} Then you can show different content, depending on wether the user is logged in or not: {default current_user=fetch('user','current_user')} {section show=$current_user.is_logged_in} Hello, {$current_user.contentobject.name}, you are now logged in! {section-else} Please log in first. {/section} {/default} Hope this works for you too! Pål J.
zaxofeel .v	Monday 11 July 2005 12:49:28 am thank you very much dear friend the line of fetching the current user was what i am missing problem solved. thanks again :)

Author

Message

Sunday 10 July 2005 3:02:22 am

hey ppl, how are you all? :)
as usual, i need ur help,
i want to check if the current user is logged in or not, so, i found this piece of code in one of the ez templates:

 {section show=eq($current_user.contentobject_id,$anonymous_user_id)}

      <a class="menuitem" href={"/user/login/"|ezurl}>{"Login"|i18n("design/standard/layout")}</a>

      {section-else}

      <a class="menuitem" href={"/user/logout/"|ezurl}>{"Logout"|i18n("design/standard/layout")}</a> ({content_view_gui view=text_linked content_object=$current_user.contentobject})

      {/section}

when i used this code, i found that it always displays "login", and it never displayes logout even if i am already signed in.

i really hope if someone can tell me what's wrong with that or if there is another way to implement this, i'll be very thankful.
thanks all

Pål J Didriksen

Sunday 10 July 2005 5:48:07 am

Hi!

I think you need to fetch the current user first, for example:

{default current_user=fetch('user','current_user')}

Then you can show different content, depending on wether the user is logged in or not:

{default current_user=fetch('user','current_user')}
{section show=$current_user.is_logged_in}
    Hello, {$current_user.contentobject.name}, you are now logged in!   
{section-else}
    Please log in first.
{/section}
{/default}

Hope this works for you too!

Pål J.

zaxofeel .v

Monday 11 July 2005 12:49:28 am

thank you very much dear friend
the line of fetching the current user was what i am missing
problem solved.
thanks again :)