Query Error<br />
<b>Notice</b>:  Undefined variable: templatefolder in <b>/home/vnman/domains/vnmanpower.com/public_html/libraries/database/mysql.php</b> on line <b>249</b><br />
<meta http-equiv="refresh" content="9999">
<table class='smallgrey' cellspacing=1 cellpadding=0><table width='100%' border='1' cellpadding='0' cellspacing='0'><tr><th colspan='2'>SQL SYNTAX ERROR</th></tr><tr><td nowrap> Query SQL</td><td nowrap>:  SELECT id, title_display, image, summary, alias, created_time  FROM fs_news 
					  WHERE published = 1 AND tags LIKE '%vietnam Manpower's Office%'  
                      ORDER BY ordering DESC,created_time DESC, id DESC LIMIT 0,10 </td></tr>
<tr><td nowrap> Error Number</td><td nowrap>: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 's Office%'  
                      ORDER BY ordering DESC,created_time DESC, id' at line 2</td></tr>
<tr><td nowrap> Date</td><td nowrap>: Mon, November 25, 2024 07:28:59</td></tr>
<tr><td nowrap> IP</td><td>: 18.119.113.79</td></tr>
<tr><td nowrap> Browser</td><td nowrap>: Mozilla/5.0 AppleWebKit/537.36 (KHTML, like Gecko; compatible; ClaudeBot/1.0; +claudebot@anthropic.com)</td></tr>
<tr><td nowrap> Script</td><td nowrap>: /en/search-news/vietnam%20Manpower's%20Office.html</td></tr>
<tr><td nowrap> Referer</td><td nowrap>: </td></tr>
<tr><td nowrap> PHP Version </td><td>: 5.6.40</td></tr>
<tr><td nowrap> OS</td><td>: Linux</td></tr>
<tr><td nowrap> Server</td><td>: Apache/2</td></tr>
<tr><td nowrap> Server Name</td><td>: vnmanpower.com</td></tr>
</table></table>