Question

我收到以下错误：

SQLSTATE[42000]: Syntax error or access violation: 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 'range join con_cat_lang as catlang join con_cat as cat left join con_frontend' at line 1

有人知道如何解决这个错误吗？我正在使用mySQL 5.5.35

原始查询应为：

    $results = $this->db->fetchAll(Con_Db :: tblprefix("" .
    "select " .
    "   concat('|_', concat(repeat('___', count(parent.idcat) - 1), catlang.name)) AS name, " .
    "   catlang.idcat as idcat, " .
    "   catlang.name as catname, " .
    "   catlang.urlname as urlname, " .
    "   cat.parentid as parentid, " .
    "   count(parent.idcat) as level, " .
    "   if (catlang.visible = 1 && (catlang.public = 1 || member.idfrontenduser IS NOT NULL), 1, 0) AS visible " .
    "from " .
    "   con_cat_tree as node " .
    "   join con_cat_tree as parent " .
    "   join con_cat_tree as range " .
    "   join con_cat_lang as catlang " .
    "   join con_cat as cat " .
    "   left join con_frontendpermissions as permission ON ((permission.item = catlang.idcatlang OR permission.item = '__GLOBAL__') AND permission.idlang = catlang.idlang AND permission.plugin ='category') " .
    "   left join con_frontendgroupmembers as member ON permission.idfrontendgroup = member.idfrontendgroup AND member.idfrontenduser = ? " .
    "where " .
    "   node.lft between parent.lft and parent.rgt " .
    "   and node.lft between range.lft and range.rgt " .
    "   and node.idcat = catlang.idcat " .
    "   and parent.tree = ? " .
    "   and catlang.idlang = ? /* language */ " .
    "   and catlang.idcat = cat.idcat " .
    "   and range.idcat = ? /* main category */ " .
    "   and catlang.idcat != range.idcat " .
    "   and catlang.visible = 1 /* category online */ " .
    "   and ( " .
    "       cat.parentid in ( " .
    "           select parent.idcat " .
    "           from " .
    "               con_cat_tree AS node, " .
    "               con_cat_tree AS parent, " .
    "               con_cat_lang AS catlang " .
    "           where " .
    "               node.lft BETWEEN parent.lft and parent.rgt " .
    "               and node.idcat = catlang.idcat " .
    "               and node.idcat = ? /* selected category */ " .
    "               and parent.tree = ? " .
    "           ) " .
    "       ) " .
    "group by node.idcat " .
    "order by node.lft" .
    ""), array (
        $this->userId,
        $this->idclient,
        $this->langId,
        $this->startId,
        $this->idcat,
        $this->idclient
    ));

Answer 1

range是reserved word。你必须逃脱它：

SELECT `reserved` FROM ...
       ^--------^--- backticks to escape

mySQL 5.5中范围的SQL错误

1 个答案: