我有以下 SQL Server 2008 查询:

SELECT T.*,Data.Value FROM [Table] T OUTER APPLY      

(SELECT TOP 1 E.Value FROM [Table2] E     
ORDER BY CASE WHEN T.TDateTime >= E.EDateTime then 1 else 2 end,
ABS(DateDiff(ss,T.TDateTime,E.EDatetime))) AS Data

这基本上得到了最后E T 中每条记录的值, 但如果记录在 T在 E 中的第一条记录之前, 然后它获得了 E 中的第一条记录.

MySQL 中的等价物是什么？

编辑

这是我的架构和数据:

DROP TABLE IF EXISTS `data`;
CREATE TABLE `data` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `DataDateTime` datetime DEFAULT NULL,
  `Value` int(11) DEFAULT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8;


LOCK TABLES `data` WRITE;
INSERT INTO `data` VALUES (1,'2012-02-01 00:00:00',1),(2,'2012-03-01 01:00:00',2),(3,'2012-04-01 02:00:00',3),(4,'2012-05-01 03:00:00',4),(5,'2012-06-01 04:00:00',5),(6,'2012-07-01 05:00:00',6),(7,'2012-08-01 06:00:00',7),(8,'2012-09-01 07:00:00',8);
UNLOCK TABLES;


DROP TABLE IF EXISTS `t`;
CREATE TABLE `t` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `TDateTime` datetime DEFAULT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8;

LOCK TABLES `t` WRITE;
INSERT INTO `t` VALUES (1,'2012-01-01 00:00:00'),(2,'2012-02-01 00:00:00'),(3,'2012-02-01 12:00:00'),(4,'2012-03-01 00:00:00'),(5,'2012-04-01 00:00:00'),(6,'2012-05-01 12:00:00'),(7,'2012-06-01 00:00:00'), (8,'2012-07-01 00:00:00');
UNLOCK TABLES;

SQLFiddle:

CREATE TABLE `data` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `DataDateTime` datetime DEFAULT NULL,
  `Value` int(11) DEFAULT NULL,
  PRIMARY KEY (`ID`)
);

INSERT INTO `data` (`DataDateTime`, `Value`) VALUES 
('2012-02-01 00:00:00',1),
('2012-03-01 01:00:00',2),
('2012-04-01 02:00:00',3),
('2012-05-01 03:00:00',4),
('2012-06-01 04:00:00',5),
('2012-07-01 05:00:00',6),
('2012-08-01 06:00:00',7),
('2012-09-01 07:00:00',8);


CREATE TABLE `t` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `TDateTime` datetime DEFAULT NULL,
  PRIMARY KEY (`ID`)
);

INSERT INTO `t` (`TDateTime`) VALUES 
('2012-01-01 00:00:00'),
('2012-02-01 00:00:00'),
('2012-02-01 12:00:00'),
('2012-03-01 00:00:00'),
('2012-04-01 00:00:00'),
('2012-05-01 12:00:00'),
('2012-06-01 00:00:00'),
('2012-07-01 00:00:00');

我想要的输出:

T.ID, T.TDateTime, Data.DataDateTime, Data.Value
1, 2012-01-01 00:00:00, 2012-02-01 00:00:00, 1
2, 2012-02-01 00:00:00, 2012-02-01 00:00:00, 1
3, 2012-02-01 12:00:00, 2012-02-01 00:00:00, 1
4, 2012-03-01 00:00:00, 2012-02-01 00:00:00, 1
5, 2012-04-01 00:00:00, 2012-03-01 01:00:00, 2
6, 2012-05-01 12:00:00, 2012-05-01 03:00:00, 4
7, 2012-06-01 00:00:00, 2012-05-01 03:00:00, 4
8, 2012-07-01 00:00:00, 2012-06-01 04:00:00, 5

最佳答案

这是我的提交:)

select *, if(Segment1Time<=ifnull(Segment2Time,Segment1Time),
             Segment1Value,
             Segment2Value) Value
from
(
  select *,
    (select DataDateTime from `data` where DataDateTime<=t.TDateTime order by DataDateTime desc limit 1) Segment1Time,
    (select Value from `data` where DataDateTime<=t.TDateTime order by DataDateTime desc limit 1) Segment1Value,
    (select DataDateTime from `data` where DataDateTime> t.TDateTime order by DataDateTime limit 1) Segment2Time,
    (select Value from `data` where DataDateTime> t.TDateTime order by DataDateTime limit 1) Segment2Value
  from `t` t
) X
order by tdatetime;

这是查询的解释。这个查询的好处是，在 data.datadatetime 上有一个索引，它与 4 个子查询是线性的，所有这些子查询都生成单个 SEEK，而不必遍历所有记录来对它们进行排名。理论上，data 表越大，效果越好。

ID  SELECT_TYPE TABLE   TYPE    POSSIBLE_KEYS   KEY KEY_LEN REF ROWS    EXTRA
1   PRIMARY <derived2>  ALL (null)  (null)  (null)  (null)  8   Using filesort
2   DERIVED t   ALL (null)  (null)  (null)  (null)  8   
6   DEPENDENT SUBQUERY  data    ALL DataDateTime    (null)  (null)  (null)  8   Using where; Using filesort
5   DEPENDENT SUBQUERY  data    index   DataDateTime    DataDateTime    9   (null)  1   Using where; Using index
4   DEPENDENT SUBQUERY  data    ALL DataDateTime    (null)  (null)  (null)  8   Using where; Using filesort
3   DEPENDENT SUBQUERY  data    index   DataDateTime    DataDateTime    9   (null)  1   Using where; Using index

关于MySQL - 按时间获取最后一个条目(如果为空则为第一个)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8529049/