mysql中GROUP_CONCAT的使用
发布时间:2020-12-11 23:16:42 所属栏目:MySql教程 来源:网络整理
导读:现在有三个表,结构如下: cate表CREATE TABLE `cate` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT COMMENT 'id',`name` char(20) DEFAULT '' COMMENT '分类名',PRIMARY KEY (`id`)) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8 COMMENT='文
现在有三个表,结构如下:
cate表 CREATE TABLE `cate` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT COMMENT 'id',`name` char(20) DEFAULT '' COMMENT '分类名',PRIMARY KEY (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8 COMMENT='文章分类表'; article表 CREATE TABLE `article` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT COMMENT 'id',`title` varchar(50) DEFAULT '',`cate_id` int(11) NOT NULL DEFAULT '0' COMMENT '分类id',PRIMARY KEY (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8 COMMENT='文章表'; article_extend表 CREATE TABLE `article_extend` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT,`article_id` int(10) unsigned DEFAULT '0' COMMENT '文章id',`name` varchar(255) DEFAULT '' COMMENT '音频,图片之类',PRIMARY KEY (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8 COMMENT='附件表';三张表数据如下: cate表: SELECT a.id AS aid,a.title AS atitle,c. NAME AS cname,ae. NAME AS aname FROM article AS a LEFT JOIN cate AS c ON a.cate_id = c.id LEFT JOIN article_extend AS ae ON a.id = ae.article_id WHERE a.id = 1;结果如下,出现了两条数据: SELECT a.id AS aid,GROUP_CONCAT(ae. NAME SEPARATOR '-') AS aname FROM article AS a LEFT JOIN cate AS c ON a.cate_id = c.id LEFT JOIN article_extend AS ae ON a.id = ae.article_id WHERE a.id = 1;结果如下: SELECT a.id AS aid,ae.allname FROM article AS a LEFT JOIN ( SELECT ae.article_id,GROUP_CONCAT(ae. NAME) AS allname FROM article_extend AS ae GROUP BY ae.article_id ) AS ae ON a.id = ae.article_id LEFT JOIN cate AS c ON a.cate_id = c.id;结果如下: (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |