列的最大值

Question

使用此脚本，我获得该列的最大值。

SELECT p.ploegnaam, MAX(k.punten) AS punten
FROM ploeg p,
     klassement k
WHERE p.id = k.ploeg

但是当我有两个具有相同值的值时，它们是最大值。我只是 得到第一个。

当该列的最大值为20且该列中有3个值为20时，如何获取所有这些值而不仅仅是该列的第一个值为20？

db script

/*Table structure for table `klassement` */

DROP TABLE IF EXISTS `klassement`;

CREATE TABLE `klassement` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `punten` int(2) NOT NULL,
  `ploeg` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `ploeg` (`ploeg`),
  CONSTRAINT `klassement_ibfk_1` FOREIGN KEY (`ploeg`) REFERENCES `ploeg` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=22 DEFAULT CHARSET=latin1;

/*Data for the table `klassement` */

insert  into `klassement`(`id`,`punten`,`ploeg`) values 

(14,7,29),

(15,7,30),

(16,1,31),

(17,0,32),

(18,0,34),

(19,1,36),

(20,0,40),

(21,0,41);

/*Table structure for table `ploeg` */

DROP TABLE IF EXISTS `ploeg`;

CREATE TABLE `ploeg` (
  `id` int(3) NOT NULL AUTO_INCREMENT,
  `Ploegnaam` varchar(50) CHARACTER SET latin1 NOT NULL,
  `Stadium` int(3) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `Stadium` (`Stadium`),
  CONSTRAINT `ploeg_ibfk_1` FOREIGN KEY (`Stadium`) REFERENCES `stadium` (`id`),
  CONSTRAINT `ploeg_ibfk_2` FOREIGN KEY (`Stadium`) REFERENCES `stadium` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=42 DEFAULT CHARSET=utf8;

/*Data for the table `ploeg` */

insert  into `ploeg`(`id`,`Ploegnaam`,`Stadium`) values 

(29,'Club Brugge',41),

(30,'Cercle Brugge',41),

(31,'België',41),

(32,'belgiê',41),

(33,'belgië',41),

(34,'belgie',41),

(35,'belgiê',41),

(36,'Mùnchen',41),

(37,'mùnchen',41),

(38,'mùnchen',41),

(39,'mùnchen',41),

(40,'mùchuo',41),

(41,'ùùùù',41);

/*Table structure for table `stadium` */

DROP TABLE IF EXISTS `stadium`;

CREATE TABLE `stadium` (
  `id` int(3) NOT NULL AUTO_INCREMENT,
  `Naam` varchar(50) CHARACTER SET latin1 NOT NULL,
  `info` varchar(240) CHARACTER SET latin1 DEFAULT NULL,
  `Adres` varchar(2048) CHARACTER SET latin1 NOT NULL,
  `afbeelding` varchar(240) CHARACTER SET latin1 NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=52 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

/*Data for the table `stadium` */

insert  into `stadium`(`id`,`Naam`,`info`,`Adres`,`afbeelding`) values 

(41,'Jan breydel','jan breydel in brugge','Jan Breydelstadion, Koning Leopold III-laan, Brugge, Belgi&euml','../uploads/volkswagen.jpg'),

(42,'jan','f','Kortrijk, België','../uploads/12042828_428221037364899_823789106058625890_n.jpg'),

(43,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(44,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(45,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(46,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(47,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(48,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(49,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(50,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg'),

(51,'jab','bebe','Leuven, België','../uploads/12417623_10206864642527072_5223843936076001877_n.jpg');

/*!40101 SET SQL_MODE=@OLD_SQL_MODE */;
/*!40014 SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS */;
/*!40014 SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS */;
/*!40111 SET SQL_NOTES=@OLD_SQL_NOTES */;

由于

Answer 1

我认为您可以使用嵌套查询：

SELECT p.ploegnaam, k.punten AS punten
FROM ploeg p inner join klassement k on p.id = k.ploeg
WHERE k.punten = (select max(punten) from klassement )

Answer 2

你可以用这个

select p.ploegnaam, k.punten
from ploeg p
inner join klassement k
    on p.id = k.ploeg
inner join (SELECT p.ploegnaam, MAX(k.punten) AS max_punten
            FROM ploeg p1
            inner join klassement k1
                on p1.id = k1.ploeg
            ) m
    on p.id = m.ploegnaam and k.punten = m.max_punten;

从现在开始尝试不使用old-style-join。