Hi Phillipe,

You almost get it right, where you need to alter it a bit where you need to add another condition if the group with that id only show the button eg:
[gist type="php"]
<?php if ($this->my->getAccess()->get('groups.allow.join') && !$group->isInviteOnly() && !$group->isMember() && !$group->isPendingMember() && !$group->isInvited() && ($group->id == 2)) { ?>
[/gist]

and if the user if already a member you can do the condition as:
[gist type="php"]
<?php if ($group->isMember()) { ?>
<?php } ?>
[/gist]
so that the user will not see the button anymore.

Thanks for your return Fadhli.
Ah, I see, it's the magic formula I missed : ($group->id == 2)

Only I get an error :

#0 Using $this when not in object context

Because in my opinion it is because we are in a module, and not in the Group.
And we must probably specify the context.
Something is missing...

An idea Fadhli ?
Thank you,
Philippe

PS.
Here is the complete content of my file : /modules/mod_easysocial_quickpost/tmpl/default.php
Display in Dashboard and Profile.

<?php

/**

* @package      EasySocial

* @copyright    Copyright (C) 2010 - 2016 Stack Ideas Sdn Bhd. All rights reserved.

* @license      GNU/GPL, see LICENSE.php

* EasySocial is free software. This version may have been modified pursuant

* to the GNU General Public License, and as distributed it includes or

* is derivative of works licensed under the GNU General Public License or

* other free or open source software licenses.

* See COPYRIGHT.php for copyright notices and details.

*/

defined('_JEXEC') or die('Unauthorized Access');

?>

<div id="es" class="mod-es mod-es-quickpost <?php echo $lib->getSuffix();?>" data-quickpost-module>



<?php if ($this->my->getAccess()->get('groups.allow.join') && !$group->isInviteOnly() && !$group->isMember() && !$group->isPendingMember() && !$group->isInvited() && ($group->id == 2)) { ?>

	<a class="btn btn-es-default-o btn-sm" href="javascript:void(0);" data-es-groups-join data-id="2" data-page-reload="1">

		<div class="o-loader o-loader--sm"></div>

		<?php echo JText::_('COM_EASYSOCIAL_GROUPS_JOIN_GROUP');?>

	</a>

<?php } ?>



	<?php echo $story->html(true);?>

</div>

Basically the
[gist type="php"]
$this->my
[/gist]
is
[gist type="php"]
ES::user();
[/gist]

Perhaps, can you try change it to
[gist type="php"]
ES::user()->getAccess()->get('groups.allow.join')
[/gist]
and see how it goes?

Thanks. There is progress but it does not work

I get this error message :

#0 Call to a member function isInviteOnly() on null

With this code :

<?php if (ES::user()->getAccess()->get('groups.allow.join') && !$group->isInviteOnly() && !$group->isMember() && !$group->isPendingMember() && !$group->isInvited() && ($group->id == 2)) { ?>

	<a class="btn btn-es-default-o btn-sm" href="javascript:void(0);" data-es-groups-join data-id="2" data-page-reload="1">

		<div class="o-loader o-loader--sm"></div>

		<?php echo JText::_('COM_EASYSOCIAL_GROUPS_JOIN_GROUP');?>

	</a>

<?php } ?>

Philippe

You need to called the group library for the $group
as example:
[gist type="php"]
$group = ES::group($groupId);
[/gist]

It was the piece that was missing from my puzzle !
It works as I want now.

Thank you very much Fadhli.
Philippe

Great Phiippe

Just for your information, I have locked and marked this thread as resolved to avoid confusions in the future. Please start a new thread if you have any other issue in the future so it will be easier for us to manage your inquiries.

Thanks for understanding and have a nice day ahead