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-   -   difficulties using blog_category.description (https://vborg.vbsupport.ru/showthread.php?t=205609)

Taragon 02-16-2009 10:14 PM

difficulties using blog_category.description
 
Hello,

Currently I'm having some difficulties using blog_category.description within a mod.
Code:

// query to get the last blog entries
$latestblogs = $db->query_read_slave("
        SELECT
                blog.blogid,
                blog.title,
                blog_text.pagetext AS pagetext,
                blog.comments_visible,
                blog.username,
                blog.dateline,
                blog.firstblogtextid,
                blog_text.blogtextid,
                blog.userid,
                blog.views,
                blog_category.description AS description
        FROM " . TABLE_PREFIX . "blog AS blog
        INNER JOIN " . TABLE_PREFIX . "blog_text AS blog_text ON (blog.firstblogtextid = blog_text.blogtextid)
        LEFT JOIN " . TABLE_PREFIX . "blog_user AS blog_user ON (blog_user.bloguserid=blog.userid)
        WHERE blog.state = 'visible'
                AND blog.pending = 0
                AND blog_user.options_guest = 3
        ORDER BY blog.blogid DESC
        LIMIT " . $vbulletin->options['lb_limit'] . "
");

while ($showblogs=$db->fetch_array($latestblogs))
{
        // strips the bbcode out of the post
        $showblogs['pagetext'] = strip_bbcode($showblogs['pagetext'], $stripquotes = true, $fast_and_dirty = false, $showlinks = true);

        // adds the cutoff ... with substr
        $showblogs['pagetext'] = substr($showblogs['pagetext'], 0, $vbulletin->options['lb_cut']) . '...';

        // convert pagetext to nice looking variable
        $counter++;
        $blog_userid = $showblogs['userid'];
        $blogid = $showblogs['blogid'];
        $blog_title = $showblogs['title'];
        $blog_pagetext = $showblogs['pagetext'];
        $blog_views = $showblogs['views'];
        $blog_comments = $showblogs['comments_visible'];
        $blog_username = $showblogs['username'];
        $blog_description = $showblogs['description'];
        $blog_dateline = vbdate($vbulletin->options['dateformat'], $showblogs['dateline']).' ';
        $blog_dateline .= vbdate($vbulletin->options['timeformat'], $showblogs['dateline']);

        eval('$latest_blogs .= "' . fetch_template('forumhome_latest_blog_bit') . '";');
}

eval('$latest_blogs = "' . fetch_template('forumhome_latest_blog') . '";');

Code:

Database error in vBulletin 3.8.0:

Invalid SQL:

        SELECT
                blog.blogid,
                blog.title,
                blog_text.pagetext AS pagetext,
                blog.comments_visible,
                blog.username,
                blog.dateline,
                blog.firstblogtextid,
                blog_text.blogtextid,
                blog.userid,
                blog.views,
                blog_category.description AS description
        FROM blog AS blog
        INNER JOIN blog_text AS blog_text ON (blog.firstblogtextid = blog_text.blogtextid)
        LEFT JOIN blog_user AS blog_user ON (blog_user.bloguserid=blog.userid)
        WHERE blog.state = 'visible'
                AND blog.pending = 0
                AND blog_user.options_guest = 3
        ORDER BY blog.blogid DESC
        LIMIT 5;

MySQL Error  : Unknown column 'blog_category.description' in 'field list'
Error Number  : 1054
Request Date  : Monday, February 16th 2009 @ 07:09:39 PM
Error Date    : Monday, February 16th 2009 @ 07:09:39 PM
Script        : http://die16-clan.com/forum.php
Referrer      :
IP Address    : 77.166.91.51
Username      : [DIE16]Taragon
Classname    : vB_Database
MySQL Version : 5.0.67-community-log

Does someone know what I might be doing wrong here?

--------------- Added [DATE]1234834398[/DATE] at [TIME]1234834398[/TIME] ---------------

never mind this.

I was looking for a way to create an header image from this, like GARS, however I didn't think about multiple categories per blog...

looking for an alternative solution now :)

Dismounted 02-17-2009 04:44 AM

You need to join the "blog_category" to the query to be able to use it.


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