我有下表:
CREATE TABLE IF NOT EXISTS `article` (
`article_id` int(11) NOT NULL,
`language_id` int(11) NOT NULL,
`category_id` int(11) NOT NULL,
`article_title` varchar(100) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
`article_body` varchar(8192) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
`meta_description` varchar(255) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
`meta_keyword` varchar(255) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
`article_main_image` varchar(255) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
`promote` bit(1) NOT NULL DEFAULT b'0',
`created` datetime NOT NULL,
`modified` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`deleted` bit(1) NOT NULL DEFAULT b'0',
`published` bit(1) NOT NULL DEFAULT b'1',
`user_id` int(11) NOT NULL,
PRIMARY KEY (`article_id`,`language_id`),
KEY `article_title` (`article_title`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
lagnuage_id:
英语= 1
阿拉伯语= 2
俄语= 3
我想select * where language_id = 1 limit by 10
,但如果他们有相同的article_id
,请将语言列加入一个。例如:
(1, 1, 0, 'qwq', 'da sdsad as da ds', 'asd sa d', 'as dsadasd as d sa', '', b'0', '2013-06-20 11:01:32', '2013-06-20 08:01:32', b'0', b'1', 0),
(1, 2, 0, '????????', '?? ??? ??? ??? ???', '?? ??? ??? ? ???? ', '??? ??? ?? ???? ?? ???? ', '', b'0', '2013-06-20 11:01:32', '2013-06-20 08:01:32', b'0', b'1', 0),
(1, 3, 0, 'rrrrrrrrrrr', 'rrrr', 'rrrrrrrrrrrrrrrrr', 'rrrrrrrrrrrrrrrrrrrrrrrrrr', '', b'0', '2013-06-20 11:01:32', '2013-06-20 08:01:32', b'0', b'1', 0),
我想选择这样的行:
(1, 1-2-3, 0, 'qwq', 'da sdsad as da ds', 'asd sa d', 'as dsadasd as d sa', '', b'0', '2013-06-20 11:01:32', '2013-06-20 08:01:32', b'0', b'1', 0),
注意语言id是如何连接的;我希望能够知道本文可用的语言。
答案 0 :(得分:0)
使用GROUP BY
和GROUP_CONCAT
我能够解决我的问题:
SELECT `article_id` , GROUP_CONCAT( `language_id` ) , ..,..,..,..
FROM `article`
GROUP BY `article_id`
LIMIT 0 , 30