mysql 基于自连接的枚举值

Question

如何使用自联接根据枚举状态获取计数

我有一个名为“student”的表，其中我使用枚举表示状态 0、1、2，其中 0 表示非活动，1 表示活动，2 表示已删除。

那是

studentid studentenrollid studentname studentstatus

1          3                 xyz          1

2          3                 xyz          2

3          8                 asda         1

4          7                 sdd          1

5          9                 asds         0

6          3                 xyz          1

我需要根据 studentrollid 唯一的状态获取计数。

studentrollid | inactive | active | deleted

    3            0           2        1

    8            0           1        0

    7            0           1        0

    9            1           0        0

Answer 1

请试试这个：

SQLFIDDLE DEMO

select studentenrollid,
sum(case when studentstatus = 1 then 1
    else 0 end) as Active, 
sum(case when studentstatus = 0 then 1
    else 0 end) as Inactive, 
sum(case when studentstatus = 2 then 1
    else 0 end) as Deleted
from demo
group by studentenrollid
;

| STUDENTENROLLID | ACTIVE | INACTIVE | DELETED |
-------------------------------------------------
|               3 |      2 |        0 |       1 |
|               7 |      1 |        0 |       0 |
|               8 |      1 |        0 |       0 |
|               9 |      0 |        1 |       0 |

Another shorter query:

select a.studentenrollid,
sum(a.studentstatus = 1) as Active, 
sum(a.studentstatus = 0) as Inactive,
sum(a.studentstatus = 2) as Deleted 
from demo a
group by a.studentenrollid
;

| STUDENTENROLLID | ACTIVE | INACTIVE | DELETED |
-------------------------------------------------
|               3 |      2 |        0 |       1 |
|               7 |      1 |        0 |       0 |
|               8 |      1 |        0 |       0 |
|               9 |      0 |        1 |       0 |

MYSQL 5.1.61 version query

---

基于 OP 在 cmets 中给出的 create statment/insert 语句：

SQLFIDDEL DEMO 2

| STUDENTENROLLID | ACTIVE | INACTIVE | DELETED |
-------------------------------------------------
|               1 |      1 |        0 |       0 |
|               2 |      1 |        0 |       0 |
|               3 |      1 |        0 |       0 |
|              41 |      9 |        1 |       1 |

Answer 2

您可以使用带有 CASE 表达式的聚合函数来透视数据：

select studentenrollid,
  count(case when studentstatus = '0' then studentid end) InActive,
  count(case when studentstatus = '1' then studentid end) Active,
  count(case when studentstatus = '2' then studentid end) Deleted
from yourtable
group by studentenrollid

见SQL Fiddle with Demo。

结果是：

| STUDENTENROLLID | INACTIVE | ACTIVE | DELETED |
-------------------------------------------------
|               3 |        0 |      2 |       1 |
|               7 |        0 |      1 |       0 |
|               8 |        0 |      1 |       0 |
|               9 |        1 |      0 |       0 |

使用您放置在 cmets 中的示例数据编辑 #1：

CREATE TABLE IF NOT EXISTS demo 
( 
  studentid int(11) NOT NULL AUTO_INCREMENT, 
  studentenrollid int(11) NOT NULL, 
  studentstatus enum('0','1','2') NOT NULL DEFAULT '1', 
  PRIMARY KEY (studentid), 
  KEY studentenrollid (studentenrollid) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=84 ;


INSERT INTO demo (studentid, studentenrollid, studentstatus) 
VALUES
(22, 41, '2'),
(23, 1,'1'),
(24, 2, '1'),
(25, 3, '1'),
(26, 41, '1'),
(29, 41, '1'),
(30, 41, '1'),
(31, 41, '1'),
(32, 41, '1'),
(33, 41, '1'),
(34, 41,'1'),
(35, 41, '1'),
(36, 41, '1'),
(37, 41, '1') 

您的查询将是：

select studentenrollid,
  count(case when studentstatus = '0' then studentid end) InActive,
  count(case when studentstatus = '1' then studentid end) Active,
  count(case when studentstatus = '2' then studentid end) Deleted
from demo
group by studentenrollid

结果（见SQL Fiddle with Demo）：

| STUDENTENROLLID | INACTIVE | ACTIVE | DELETED |
-------------------------------------------------
|               1 |        0 |      1 |       0 |
|               2 |        0 |      1 |       0 |
|               3 |        0 |      1 |       0 |
|              41 |        0 |     10 |       1 |

请注意，您只会得到一个已删除的计数，因为这是您在数据库中拥有的唯一值（至少在示例数据中）。