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Current site access... Where is the template var?

Current site access... Where is the template var?

Tuesday 24 June 2003 3:51:54 pm - 8 replies

Author Message

Selmah Maxim

Wednesday 25 June 2003 1:17:22 am

try this :

{ezini('SiteSettings','SiteURL')}

Björn Dieding@xrow.de

Wednesday 25 June 2003 6:32:12 am

{ezini('SiteSettings','SiteURL')}
gives
www:92
Thats not it.... That's the simple siteurl.

my access types are called "english" and "german"

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Bård Farstad

Thursday 10 July 2003 5:30:30 am

And the answer is: {$access_type.name}

A bit late though ;)

--bård

Documentation: http://ez.no/doc

Jan Komárek

Monday 09 August 2010 9:58:57 am

Its functional only in pagelyout for me. Is any way how to do this in non layout template?

Kévin S.

Tuesday 10 August 2010 2:41:37 am

You can use template parameters :

{include uri='design:template_name.tpl' accessType=$access_type.name}

Then in your 'template_name.tpl' you just have to call the given parameter as a variable :

{$accessType|attribute(show, 1)}

It works for view templates as well as system templates.

Jan Komárek

Tuesday 07 September 2010 10:14:00 am

It works for view templates as well as system templates.

 I am not sure what you mean by system template. But I still have no idea how can I get $access_type in for example content/view/full/article.tpl.

Thank you very much.

Abdelkader RHOUATI

Tuesday 07 September 2010 5:32:44 pm

Hi,

I had the same problem today, for solution i used a template operators "current_siteaccess" available in the extension swark (http://ez.no/developer/contribs/template_plugins/swark). In addition to this operator, the extension provides severals PHP's functions that are not in the list of defaults operators of ezpublish. I use it in almost all projects.

And most important "current_siteaccess" works in every template ;).

TKS.

Abdelkader RHOUATI

Blog (french) : http://arhouati.com
----
Extension arh_jdebug : EzDebug using jquery

Jan Komárek

Friday 10 September 2010 8:41:41 am

Thank you, thats it!

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