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[Paid work] "Sign in/Sign Out" link outside Vanilla
Hi all, my last try didn't go so well so I've simplified what I need and am here to try again.
What I need is for the "Not signed in (Sign In)" link to appear outside a Vanilla installation. And if logged in, the "Signed in as wohoo (Sign Out)" link. No login forms, just that simple line of text that appears on the top-right side of Vanilla.
I have a Wordpress 2.5 site and I'd really like to get this done without merging databases.
If it makes a difference, this is how I have things set up:
WP: www.domain.com
Vanilla: www.domain.com/forums/
Figuring out how to do this is the only major thing standing in my way from getting my site and Vanilla exactly the way I'd like them to be, so I'd really appreciate any help. If anyone can give me a hand with this I'll gladly send a $25 their way via PayPal.
Thanks!
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Comments
1) Check "conf/settings.php"--- does $Configuration['COOKIE_PATH'] = '/forums/' ? Change it to '/'.
2) Add this where appropriate---
if (array_key_exists('lussumocookieone', $_COOKIE) && array_key_exists('lussumocookietwo', $_COOKIE)) { // User is signed IN } else { // User is signed OUT }
Want me to try to do the whole thing?
One other thing, you *might* have to place the code before any output (setting cookies works that way), but I don't think that matters in this case.
Point (2): By "add this where appropriate" you mean what exactly? Where is appropriate, in my WP theme header? And what do I wrap that code in if at all? <?php ?>, ? N00b here, remember
And what then, add this to where I want the "Sign in/out" to actually appear?:
<?php echo '<div id="Session" style="background:#000;height:23px" width="982">'; if ($this->Context->Session->UserID > 0) { echo str_replace('//1', $this->Context->Session->User->Name, $this->Context->GetDefinition('SignedInAsX')).' (<a href="'.$this->Context->Configuration['SIGNOUT_URL'].'">'.$this->Context->GetDefinition('SignOut').'</a>)'; } echo '</div>'; ?>