CREATE TABLE `cdb_labels` (
? `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
? `type` int(11) NOT NULL DEFAULT '0' COMMENT '標(biāo)簽類型：1喜歡異性類型,2擅長(zhǎng)話題',
? `content` varchar(255) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL COMMENT '標(biāo)簽內(nèi)容',
? PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=57 DEFAULT CHARSET=utf8 COMMENT='標(biāo)簽內(nèi)容';

所有數(shù)據(jù)：

SELECT * FROM cdb_labels WHERE type = 1;

這里寫圖片描述

解決思路

1.對(duì)滿足初次查詢的數(shù)據(jù)賦予一個(gè)自增列b

 select id,type,content,(@b:=@b+1) as b from cdb_labels a,(SELECT @b := 0) tmp_b where type=1;

這里寫圖片描述

2.用自增的id減去自增列b

select id,type,content,( id-(@b:=@b+1) ) as c from cdb_labels a,(SELECT @b := 0) tmp_b where type=1;

這里寫圖片描述

3.對(duì)等差列c分組, 并將分組的id組裝起來(lái)

select count(id),GROUP_CONCAT(id) from ( 
    select id,( id-(@b:=@b+1) ) as c from cdb_labels a,(SELECT @b := 0) tmp_b where type=1 
) as d GROUP BY c;

這里寫圖片描述

注：為了方便區(qū)分，這里查詢分組成員要大于5（也就是連續(xù)出現(xiàn)超過(guò)5次的記錄）:

select if( count(id)>5 ,GROUP_CONCAT(id),null) e from (?
? ? select id,( id-(@b:=@b+1) ) as c from cdb_labels a,(SELECT @b := 0) tmp_b where type=1?
) as d GROUP BY c;

那么得到的數(shù)據(jù)只有：9,10,11,12,13,14,15

這里寫圖片描述

4.根據(jù)組裝的id去找數(shù)據(jù)

select id,type,content from cdb_labels,(
    select if( count(id)>5 ,GROUP_CONCAT(id),null) e from ( 
        select id,( id-(@b:=@b+1) ) as c from cdb_labels a,(SELECT @b := 0) tmp_b where type=1 
    ) as d GROUP BY c
) as f where f.e is not null and FIND_IN_SET(id , f.e);

這里寫圖片描述