joomla 2.5类别正在崩溃

时间:2014-02-12 22:11:02

标签: joomla crash categories

在我的托管公司升级服务器后,我对这些类别有一个非常奇怪的问题。每隔一天类别崩溃,当单击链接的菜单项时,它会显示以下错误。你能帮帮忙吗?谢谢。 (http://myxolargos.gr

  

错误:500

     

表'/tmp/#sql_1325_1.MYI'的密钥文件不正确;尝试修复它   SQL = SELECT a.id,a.title,a.alias,a.title_alias,a.introtext,   a.checked_out,a.checked_out_time,a.catid,a.created,a.created_by,   a.created_by_alias,CASE WHEN a.modified = 0那么a.created ELSE   a.modified END as modified,a.modified_by,uam.name as   modified_by_name,CASE WHEN.publish_up = 0那么a.created ELSE   a.publish_up END为publish_up,a.publish_down,a.images,a.urls,   a.attribs,a.metadata,a.metakey,a.metadesc,a.access,a.hits,   a.xreference,a.featured,LENGTH(a.fulltext)AS readmore,CASE WHEN   badcats.id不为null那么0 ELSE a.state END AS state,c.title AS   category_title,c.path AS category_route,c.access AS category_access,   c.alias AS category_alias,CASE WHEN a.created_by_alias> ' ' 然后   a.created_by_alias ELSE ua.name END作者,ua.email AS   author_email,(SELECT MAX(contact.id)AS id FROM   myxolarg_contact_details AS contact WHERE contact.published = 1 AND   contact.user_id = a.created_by)as contactid,parent.title as   parent_title,parent.id为parent_id,parent.path为parent_route,   parent.alias作为parent_alias,ROUND(v.rating_sum / v.rating_count,0)   AS评级,v.rating_count为rating_count,c.published,CASE WHEN   badcats.id为null然后c.published ELSE 0 END作为parents_published   来自myxolarg_content AS LEFT JOIN myxolarg_content_frontpage AS fp   ON fp.content_id = a.id LEFT JOIN myxolarg_categories AS c ON c.id =   a.catid LEFT JOIN myxolarg_users AS ua ON ua.id = a.created_by LEFT   加入myxolarg_users AS uam ON uam.id = a.modified_by LEFT JOIN   myxolarg_categories as parent ON parent.id = c.parent_id LEFT JOIN   myxolarg_content_rating AS v ON a.id = v.content_id LEFT OUTER JOIN   (选择cat.id作为id来自myxolarg_categories AS cat JOIN   myxolarg_categories AS parent on cat.lft BETWEEN parent.lft AND   parent.rgt WHERE parent.extension ='com_content'和parent.published   != 1 GROUP BY cat.id)AS badcats ON badcats.id = c.id WHERE a.access   IN(1,1)和c.access IN(1,1)以及当badcats.id为空时的情况   a.state ELSE 0 END = 1 AND(a.catid = 109 OR a.catid IN(SELECT   sub.id来自myxolarg_categories as sub INNER JOIN myxolarg_categories   如此ON sub.lft> this.lft AND sub.rgt< this.rgt WHERE this.id =   109))AND(a.publish_up ='0000-00-00 00:00:00'或a.publish_up< =   '2014-02-12 21:58:27')AND(a.publish_down ='0000-00-00 00:00:00'或   a.publish_down> ='2014-02-12 21:58:27')订购c.lft,CASE WHEN   a.publish_up = 0那么a.created ELSE a.publish_up END DESC,a.created   LIMIT 0,11

1 个答案:

答案 0 :(得分:0)

检查您的托管帐户是否空间不足。

How do you fix a MySQL "Incorrect key file" error when you can't repair the table?表示他们的解决方案是确保tmp文件夹有足够的空间来执行repai。

如果您的托管公司刚刚更改了某些内容,那么他​​们可能已经做了一些会影响您的MYSQL的tmp文件夹位置。