XF 2.2 Show a link conditionally with html

[thunderup]

Is there any way to make something like this work within the html body of a page

<xf:if is="!$xf.visitor.user_id">
<li><a href="account">Manage Account</a></li>
<xf:else />
<li><a href="#">Create Account</a></li>
</xf:if>

 

[Paul B]

<xf:if is="$xf.visitor.user_id">
    <li><a href="{{ link('account') }}">Manage Account</a></li>
<xf:else />
    <li><a href="{{ link('register') }}">Create Account</a></li>
</xf:if>