在MySQL中实现Rank高级排名函数
前言
MySQL中没有Rank排名函数,当我们需要查询排名时,只能使用MySQL数据库中的基本查询语句来查询排名,下面有3种具体的实现。(ps:非特殊情况,实际开发尽量使用Redis的sortset去实现排行。)
用例表
接下来的排行实现都采用这个表的数据举例。(ps:建议直接跳过,只是方便测试而已。)
SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;
DROP TABLE IF EXISTS `user`;
CREATE TABLE `user` (
`id` int(0) NOT NULL AUTO_INCREMENT,
`name` varchar(25) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NULL DEFAULT NULL,
`age` int(0) NULL DEFAULT NULL,
PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;
INSERT INTO `user` VALUES (1, 'a', 0);
INSERT INTO `user` VALUES (2, 'b', 41);
INSERT INTO `user` VALUES (3, 'c', 27);
INSERT INTO `user` VALUES (4, 'd', 9);
INSERT INTO `user` VALUES (5, 'e', 26);
INSERT INTO `user` VALUES (6, 'f', 43);
INSERT INTO `user` VALUES (7, 'g', 16);
INSERT INTO `user` VALUES (8, 'h', 10);
INSERT INTO `user` VALUES (9, 'i', 5);
INSERT INTO `user` VALUES (10, 'j', 53);
INSERT INTO `user` VALUES (11, 'k', 12);
INSERT INTO `user` VALUES (12, 'l', 21);
INSERT INTO `user` VALUES (13, 'm', 8);
INSERT INTO `user` VALUES (14, 'n', 39);
INSERT INTO `user` VALUES (15, 'o', 53);
INSERT INTO `user` VALUES (16, 'p', 5);
INSERT INTO `user` VALUES (17, 'q', 60);
INSERT INTO `user` VALUES (18, 'r', 44);
INSERT INTO `user` VALUES (19, 's', 41);
INSERT INTO `user` VALUES (20, 't', 13);
INSERT INTO `user` VALUES (21, 'u', 2);
INSERT INTO `user` VALUES (22, 'v', 32);
INSERT INTO `user` VALUES (23, 'w', 32);
INSERT INTO `user` VALUES (24, 'x', 4);
INSERT INTO `user` VALUES (25, 'y', 53);
INSERT INTO `user` VALUES (26, 'z', 37);
SET FOREIGN_KEY_CHECKS = 1;
1.在MySQL中实现普通排名
SELECT id , name , age , @rank := @rank + 1 AS ageRank
FROM user , ( SELECT @rank := 0 ) s
ORDER BY age desc
2.在MySQL中实现并列连续序号排名
SELECT id , name , age ,
CASE
WHEN @ageValue = age THEN
@rank
WHEN @ageValue := age THEN
@rank := @rank + 1
END AS ageRank
FROM user , ( SELECT @rank := 0 , @ageValue := NULL ) s
ORDER BY age desc
3.在MySQL中实现并列非连续序号排名
SELECT id , name , age , ageRank
FROM (
SELECT id , name , age ,
@rank := IF( @ageValue = age ,@rank ,@realRank ) AS ageRank ,
@realRank := @realRank + 1 realRank,
@ageValue := age ageValue
FROM user , ( SELECT @rank := 0 , @ageValue := NULL , @realRank := 1) s
ORDER BY age desc
)t