Question

我有以下表

userid | Field1 | Field2 | Field3
1      | S      | M      | M
2      | M      | A      | B

现在我有一套（S，M，A）。

我想给2分。 Field1，Field3，Fields这三个包含给定集合中的任何一个值。如果任何一个Field匹配，则该点将为6.

例如输出应为

userid | Field1 | Field2 | Field3 | points
1      | S      | M      | M      | 6
2      | M      | A      | B      | 4

我使用了Followig Query但它没有返回任何值

SELECT @hqp,
       UserId 
  FROM (SELECT @hqp := 0, 
               IF( Field1 IN ('S','M','A'), @hqp := @hqp +2, @hqp),
               IF( Field2 IN ('S','M','A'), @hqp := @hqp +2, @hqp),
               IF( Field3 IN ('S','M','A'), @hqp = @hqp +2, @hqp),
               @prev:=A.user_id as UserId 
          FROM users as A 
                 JOIN (SELECT @hqp := 0, @prev:=0) as B) as M

但它为每一行返回@hpq为0。任何人都可以提供帮助。

Answer 1

尝试这种方式：

SELECT userid,
       Field1,
       Field2,
       Field3,
       (IF(Field1 IN ('S','M','A'),2,0)+
        IF(Field2 IN ('S','M','A'),2,0)+
        IF(Field3 IN ('S','M','A'),2,0) ) points
  FROM users

在此处查看：http://sqlfiddle.com/#!9/322410/1

Answer 2

这种问题往往是设计不良的症状，所以看看正常化。

但是，以下内容将起作用......

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(userid INT NOT NULL PRIMARY KEY
,Field1 CHAR(1) NOT NULL
,Field2 CHAR(1) NOT NULL
,Field3 CHAR(1) NOT NULL
);

INSERT INTO my_table VALUES
(1,'S','M','M'),
(2,'M','A','B');

SELECT userid
     , SUM(f IN ('S','M','A'))*2 points 
  FROM
     ( SELECT userid,field1 f  FROM my_table UNION ALL SELECT userid,field2 FROM my_table UNION ALL SELECT userid,field3 FROM my_table ) x
 GROUP 
    BY userid;
+--------+--------+
| userid | points |
+--------+--------+
|      1 |      6 |
|      2 |      4 |
+--------+--------+

基于匹配的Mysql计算

2 个答案: