John Davidson

Message:

I need a way to get the current logged in user object from Fluid. I need it in Partials and Layouts, not only Templates, so I can't do this by getting it as an argument from a Controller class.

I already know how to do this from Typoscript, e.g. to get his/her name:

TSFE:fe_user|user|name

To get it from Php I use my own static function from my own extended FrontendUser class: User, called User::getLogged(). Could I call this function from a Partial? Or is there any other method?

Thanks in advance!

---

My Answer:
In Fluid, you can access PHP functions and objects by using the {namespace} syntax. You can create a custom namespace in your Fluid template to access your User class and its methods.

Here's an example of how you can access the User::getLogged() method in a Fluid template:

1. Define a custom namespace in your Fluid template:

html
{namespace user=Vendor\ExtensionName\ViewHelpers}


2. Create a ViewHelper class that calls the User::getLogged() method:

php
namespace Vendor\ExtensionName\ViewHelpers;

use TYPO3\CMS\Fluid\Core\ViewHelper\AbstractViewHelper;

class GetLoggedUserViewHelper extends AbstractViewHelper
{
public function render()
{
return \Vendor\ExtensionName\Domain\Model\User::getLogged();
}
}


3. Use the custom ViewHelper in your Fluid template to get the current logged-in user:

html
{user:getLogged()}


This way, you can access the current logged-in user object in your Partials and Layouts without passing it as an argument from a Controller class.