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  #1  
Old 01-29-2003, 12:37 PM
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USODJA USODJA is offline
 
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Default Using Titles in Showgroups

Is there a way to show a persons title after their name on the staff page.

Like Coordinators, CEO, yada yada...

Thanks
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  #2  
Old 01-30-2003, 12:38 PM
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should be possible

open showgroups.php
find:
PHP Code:
$users $DB_site->query("
    SELECT
        
$locationfieldselect usergroup.title, user.username, user.userid, user.invisible, user.receivepm,
        user.usergroupid, user.lastactivity, user.lastvisit
        FROM usergroup
        LEFT JOIN user ON (usergroup.usergroupid = user.usergroupid)
        LEFT JOIN userfield ON (userfield.userid = user.userid)
        WHERE usergroup.showgroup = 1
    "
); 
and change it to:
PHP Code:
$users $DB_site->query("
    SELECT
        
$locationfieldselect usergroup.title, user.title AS usertitle, user.username, user.userid, user.invisible, user.receivepm,
        user.usergroupid, user.lastactivity, user.lastvisit
        FROM usergroup
        LEFT JOIN user ON (usergroup.usergroupid = user.usergroupid)
        LEFT JOIN userfield ON (userfield.userid = user.userid)
        WHERE usergroup.showgroup = 1
    "
); 
then find:
PHP Code:
$users $DB_site->query("
    SELECT
        
$locationfieldselect forum.forumid, forum.title AS forumtitle,
        user.username, user.userid, user.invisible, user.receivepm, user.lastactivity, user.lastvisit
        FROM moderator
        LEFT JOIN user ON (user.userid = moderator.userid)
        LEFT JOIN forum ON (forum.forumid = moderator.forumid)
        LEFT JOIN userfield ON (userfield.userid = user.userid)
    WHERE forum.active = 1
    ORDER BY user.username ASC, forum.displayorder ASC
    "
); 
and change to:
PHP Code:
$users $DB_site->query("
    SELECT
        
$locationfieldselect forum.forumid, forum.title AS forumtitle,
        user.title AS usertitle, user.username, user.userid, user.invisible, user.receivepm, user.lastactivity, user.lastvisit
        FROM moderator
        LEFT JOIN user ON (user.userid = moderator.userid)
        LEFT JOIN forum ON (forum.forumid = moderator.forumid)
        LEFT JOIN userfield ON (userfield.userid = user.userid)
    WHERE forum.active = 1
    ORDER BY user.username ASC, forum.displayorder ASC
    "
); 
then you should be able to use $user[usertitle] in the showgroupbit templates
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  #3  
Old 01-30-2003, 04:47 PM
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USODJA USODJA is offline
 
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I get this error:

Database error in vBulletin 2.2.9:

Invalid SQL:
SELECT
usergroup.title, user.title AS usertitle, user.username, user.userid, user.invisible, user.receivepm,
user.usergroupid, user.lastactivity, user.lastvisit
FROM usergroup
LEFT JOIN user ON (usergroup.usergroupid = user.usergroupid)
LEFT JOIN userfield ON (userfield.userid = user.userid)
WHERE usergroup.showgroup = 1

mysql error: Unknown column 'user.title' in 'field list'

mysql error number: 1054
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  #4  
Old 02-01-2003, 02:34 AM
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USODJA USODJA is offline
 
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  #5  
Old 02-01-2003, 09:55 AM
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replace user.title with user.usertitle
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  #6  
Old 02-01-2003, 01:15 PM
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Default

Thanks it worked like a charm, I run an association site, so they were all bugging me to list the members with titles on a page, and I knew modifiying this one would do the trick....

Thanks again!
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  #7  
Old 02-02-2003, 12:07 PM
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you're welcome
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  #8  
Old 09-24-2006, 03:53 PM
blonboy blonboy is offline
 
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I would like to be able to use this modification as well, but it seems the code has changed substantialy since this posting. Anyone willing to update what would be the necessary changes? Thanks In Advance.
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  #9  
Old 09-29-2006, 07:45 AM
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now it seems like it looks like this:
PHP Code:
// get usergroups who should be displayed on showgroups
// Scans too many rows. Usergroup Rows * User Rows
$users $db->query_read_slave("
    SELECT user.*, usergroup.usergroupid, usergroup.title, user.options, usertextfield.*, userfield.*,
    IF(displaygroupid=0, user.usergroupid, displaygroupid) AS displaygroupid, infractiongroupid
    FROM " 
TABLE_PREFIX "usergroup AS usergroup
    LEFT JOIN " 
TABLE_PREFIX "user AS user ON(user.usergroupid = usergroup.usergroupid OR FIND_IN_SET(usergroup.usergroupid, user.membergroupids))
    LEFT JOIN " 
TABLE_PREFIX "userfield AS userfield ON(userfield.userid = user.userid)
    LEFT JOIN " 
TABLE_PREFIX "usertextfield AS usertextfield ON(usertextfield.userid=user.userid)
    WHERE (usergroup.genericoptions & " 
$vbulletin->bf_ugp_genericoptions['showgroup'] . ")
"
); 
What to do now?

EDIT:
Why even go that way? You could just go to the showgroups_usergroupbit template in templatemanager.. then you could just copy the usertitle code from postbit and paste it there?
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  #10  
Old 01-18-2007, 09:19 PM
sinucello sinucello is offline
 
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Hi,

Quote:
EDIT:
Why even go that way? You could just go to the showgroups_usergroupbit template in templatemanager.. then you could just copy the usertitle code from postbit and paste it there?
ASFAIK you can only display in your template what is selected in the code. So pasting the template-parts there won`t help if you don`t extract the necessary data in the php-files using that template.

Ciao,
Sacha
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