@mysql查询最大连续天数 sql
SELECT
uid,
min(date) AS mindate,
max(date) AS maxdate,
max(countday) countday,
group_concat(date) dates
FROM(
SELECT uid,date,@countday := ( CASE WHEN ( DATEDIFF(date ,@last_date) = 1) THEN (@countday + 1) ELSE 1 END) AS countday,
@last_date := date
FROM(SELECT uid,dateFROM test_demo ORDER BY uid,date) AS t1,
(select @countday:=0,@last_date:=’’) as t2
) AS t3 GROUP BY uid
SELECT DATEDIFF(‘2015-09-02’,‘2015-09-01’)
分解sql
SELECT uid,dateFROM test_demo ORDER BY uid,date 排序查出id,日期
@countday @last_date 这种是变量声明
:= 赋值
@countday := ( CASE WHEN ( DATEDIFF(date ,@last_date) = 1) THEN (@countday + 1) ELSE 1 END
DATEDIFF(d1,d2) 计算两个时间的时间差值
当两个时间差值为1的时候,则为连续天数，
@countday + 1 变量加一，否则为1 结束
@last_date := date 将data赋值变量@last_date
所以每次比较的都是当前行data和上一行data比较。
(select @countday:=0,@last_date:=’’) as 赋值变量
测试数据
CREATE TABLE test_demo (
id int(11) NOT NULL AUTO_INCREMENT,
uid int(11) NOT NULL DEFAULT ‘0’,
date date NOT NULL,
PRIMARY KEY (id),
UNIQUE KEY idx_ud (uid,date)
) ENGINE=InnoDB AUTO_INCREMENT=23 DEFAULT CHARSET=latin1
INSERT INTO test_demo (id, uid, date) VALUES
(1, 1, ‘20111-011-01’),
(2, 1, ‘20111-011-02’),
(17, 1, ‘20111-011-03’),
(18, 1, ‘20111-011-04’),
(3, 2, ‘20111-011-01’),
(40, 2, ‘20111-011-02’),
(118, 2, ‘20111-011-03’),
(20, 2, ‘20111-011-04’),
(11, 3, ‘20111-011-01’),
(6, 3, ‘20111-011-02’),
(21, 3, ‘20111-011-03’),
(22, 3, ‘20111-011-04’),
(75, 4, ‘20111-011-01’),
(87, 4, ‘20111-011-02’),
(101, 11, ‘20111-011-01’),
(10, 11, ‘20111-011-02’),
(11, 6, ‘20111-011-01’),
(12, 6, ‘20111-011-02’),
(13, 7, ‘20111-011-01’),
(94, 7, ‘20111-011-02’),
(18, 8, ‘20111-011-01’),
(156, 8, ‘20111-011-02’);