Reputation:
it's shown if you upload resource.
Reputation:
Yes, I understand this, but how can I deactivate it? Is there a way? Thank you in advance.
Style properties all of them can be activated and deactivated
Reputation:
I can't understand your answer, where can I disable this option? in property style? Where?
Reputation:
Reputation:
I deactivated, but I still can't see any changes, should I rebuild the cache?
Is it possible that we talk about different things?
You ask how to deactivate this field, right?
Is it still shown?
Or do you want to see that field? And for the user it is not shown?
Reputation:
I want to see that field!
ok did the user uploaded any resource?
does the user have the rights to see resources?
Reputation:
I want to see the "0" resource field, the default field, what can I do?
<xf:if is="$xf.visitor.canViewResources()">
<dl class="pairs pairs--rows pairs--rows--centered menu-fauxLinkRow">
<dt>{{ phrase('xfrm_resources') }}</dt>
<dd>
<a href="{{ link('resources/authors', $user) }}" class="menu-fauxLinkRow-linkRow u-concealed">
{$user.xfrm_resource_count|number}
</a>
</dd>
</dl>
</xf:if>
$0
You can try to change the the template modification I've mentioned and remove the && $user.xfrm_resource_count like this:
PHP:<xf:if is="$xf.visitor.canViewResources()"> <dl class="pairs pairs--rows pairs--rows--centered menu-fauxLinkRow"> <dt>{{ phrase('xfrm_resources') }}</dt> <dd> <a href="{{ link('resources/authors', $user) }}" class="menu-fauxLinkRow-linkRow u-concealed"> {$user.xfrm_resource_count|number} </a> </dd> </dl> </xf:if> $0
But I have not tested it.
Reputation:
You can try to change the the template modification I've mentioned and remove the && $user.xfrm_resource_count like this:
PHP:<xf:if is="$xf.visitor.canViewResources()"> <dl class="pairs pairs--rows pairs--rows--centered menu-fauxLinkRow"> <dt>{{ phrase('xfrm_resources') }}</dt> <dd> <a href="{{ link('resources/authors', $user) }}" class="menu-fauxLinkRow-linkRow u-concealed"> {$user.xfrm_resource_count|number} </a> </dd> </dl> </xf:if> $0
But I have not tested it.