Last thread with last post icon instead of clear image

(The creator of the original hack is DR2000; I have since then modified it further)

Here's the hack in a handy dandy text file:

Hack:lastthreadlastposticon.txt

It is incredibly simple...you need only edit one file and one template! It can be done in a couple minutes!

###Features###

- Shows the thread name of the last active post on the front page
- Shows the lastpost.gif icon for the theme you're using, instead of either A, using a clear (invisible) image, or using the default lastpost.gif image for the lastpost icon
- If the subject is long enough to start stretching the field on the index page, it'll show a shrinked version (i.e. "I need help with version.....).

Note: This has only been tested on version 2.2.1!

To see it in action, just visit www.ConsoleBoards.com !

[Lesane]

[QUOTE]Originally posted by Lesane
hmm.. somehow the forum dont shows \ \ in the code.

Epic, change in the select code in index.php this:

PHP Code:

THEN '' 


to:

THEN \ ' \ '

(without spaces)

[epic]

I have a db error when use THEN \ ' \ '
no other hacks!

[Island Roots]

[QUOTE]Originally posted by epic
I have a db error when use THEN \ ' \ '
no other hacks!

[epic]

My server is php 4.0.4pl1 and MySql3.22.32
My vbb is 2.2.1
When i use this line

PHP Code:

$forums=$DB_site->query('SELECT forum.*, CASE WHEN icon.iconpath IS NULL THEN \'\' ELSE icon.iconpath END as icon, thread.title as thtitle FROM forum left outer join thread on thread.forumid = forum.forumid and thread.lastpost >= (forum.lastpost-30) LEFT OUTER JOIN icon ON icon.iconid=thread.iconid WHERE displayorder<>0 AND active=1 ORDER BY parentid,displayorder'); 


generate a Database error,like this

PHP Code:

Database error in vBulletin 2.2.1:

Invalid SQL: SELECT forum.*, CASE WHEN icon.iconpath IS NULL THEN '' ELSE icon.iconpath END as icon, thread.title as thtitle FROM forum left outer join thread on thread.forumid = forum.forumid and thread.lastpost >= (forum.lastpost-30) LEFT OUTER JOIN icon ON icon.iconid=thread.iconid WHERE displayorder<>0 AND active=1 ORDER BY parentid,displayorder
mysql error: You have an error in your SQL syntax near 'icon.iconpath IS NULL THEN '' ELSE icon.iconpath END as icon, thread.title as th' at line 1 


When i use this line

PHP Code:

$forums=$DB_site->query('SELECT forum.*, CASE WHEN icon.iconpath IS NULL THEN '' ELSE icon.iconpath END as icon, thread.title as thtitle FROM forum left outer join thread on thread.forumid = forum.forumid and thread.lastpost >= (forum.lastpost-30) LEFT OUTER JOIN icon ON icon.iconid=thread.iconid WHERE displayorder<>0 AND active=1 ORDER BY parentid,displayorder'); 


generate a error,info is
Parse error: parse error in /home/vb/index.php on line 92
Who can help me?

[Island Roots]

<i>Make sure</i> you didn't overwrite any other code besides the one you were suppose to edit. If that doesn't work, then I can't help you. It has worked perfectly on all other boards.

[trilOByte]

Getting the same parse error here too! Wierd!

[nuno]

find (index.php):

//Forum info
$forums=$DB_site->query('SELECT * FROM forum WHERE displayorder<>0 AND active=1 ORDER BY parentid,displayorder');

replace it with:

//Forum info
$forums=$DB_site->query('SELECT forum.*, CASE WHEN icon.iconpath IS NULL THEN \'\' ELSE icon.iconpath END as icon, thread.title as thtitle FROM forum left outer join thread on thread.forumid = forum.forumid and thread.lastpost >= (forum.lastpost-30) LEFT OUTER JOIN icon ON icon.iconid=thread.iconid WHERE displayorder<>0 AND active=1 ORDER BY parentid,displayorder');

[trilOByte]

Thanks man, that worked a treat