MySQL计数行

Question

如何根据其内容计算行数？ 假设我有台这样的

[表]

ID_COMPANY | NAME 
----------------------------- 
A1   | COMPANY A 

[表B]

ID_COMPANY | USER  | TYPE 
-------------------------------------- 
A1   | USER A | MANAGER 
A1   | USER B | DEPT001 
A1   | USER C | CUSTOMR 
A1   | USER D | DEPT002 
A1   | USER E | CUSTOMR 

我怎样才能得到这样的结果？

ID_COMPANY | NAME | TOTAL_MANAGER | TOTAL_STAFF_DEPT | TOTAL_CUST 
---------------------------------------------------------------------------- 
A1   | COMPANY A |    1 |    2 |   1 

THX家伙

Answer 1

SELECT 
    `table_a`.`ID_COMPANY`, 
    `NAME`, 
    SUM(IF(`TYPE` = 'MANAGER', 1, 0)) AS `TOTAL_MANAGER`, 
    SUM(IF(`TYPE` LIKE 'DEPT%', 1, 0)) AS `TOTAL_STAFF_DEPT`, 
    SUM(IF(`TYPE` = 'CUSTOMR', 1, 0)) AS `TOTAL_CUST` 
FROM `table_a` 
JOIN `table_b` 
USING (`ID_COMPANY`) 
GROUP BY `table_a`.`ID_COMPANY` 

SUM s的标准可能需要调整，因为我不完全明白你想要达到的目标。

Answer 2

使用子查询和统计，从他们的结果。

在有缺陷的“伪SQL”：

select ID_COMPANY, NAME, 
count(select * from b where type like "MAN*) as "TOTAL_MANAGER", 
count(select * from b where type like "DEPT*") as "TOTAL_STAFF_DEPT", 
count(select * from b where type like "CUST*") as "TOTAL_CUST" 

，当我说有缺陷的，我的意思是我没有尝试过这一点，我只是试图让思想跨越，而不是给你的东西只需复制&粘贴。

Answer 3

在Matthew的回复中扩展我建议你使用UNION和GROUP BY。如：

SELECT ID_COMPANY, NAME, COUNT(USER) AS TOTAL_MANAGER FROM TABLE_B WHERE TYPE LIKE 'MANAGER' GROUP BY ID_COMPANY 

您需要将这些结果合并得到一个结果集。

Answer 4

喜欢的东西：

SELECT 
    ID_COMPANY, 
    NAME, 
    (SELECT COUNT(ID_COMPANY) FROM table_b WHERE ID_COMPANY = table_a.ID_Company and TYPE = 'MANAGER') as TOTAL_MANAGER, 
    (SELECT COUNT(ID_COMPANY) FROM table_b WHERE ID_COMPANY = table_a.ID_Company and TYPE = 'DEPT001') as DEPT001C, 
    (SELECT COUNT(ID_COMPANY) FROM table_b WHERE ID_COMPANY = table_a.ID_Company and TYPE = 'DEPT002') as DEPT002C, 
FROM table_a 
GROUP BY ID_COMPANY 

Answer 5

...这是一个JOIN方法：

SELECT 
    a.id_company, 
    a.name, 
    mgr.cnt AS total_manager, 
    dept.cnt AS total_staff_dept, 
    cust.cnt AS total_cust 
FROM 
    a 
    JOIN 
    (SELECT id_company, COUNT(*) AS cnt 
    FROM b WHERE type = 'MANAGER' GROUP BY id_company) mgr 
    ON a.id_company = mgr.id_company 
    JOIN 
    (SELECT id_company, COUNT(*) AS cnt 
    FROM b WHERE type LIKE 'DEPT%' GROUP BY id_company) dept 
    ON a.id_company = dept.id_company 
    JOIN 
    (SELECT id_company, COUNT(*) AS cnt 
    FROM b WHERE type = 'CUSTOMR' GROUP BY id_company) cust; 

......这给了我（假设没有更多的记录比你显示）：

+------------+------+---------------+------------------+------------+ 
| id_company | name | total_manager | total_staff_dept | total_cust | 
+------------+------+---------------+------------------+------------+ 
| A1   | foo |    1 |    2 |   2 | 
+------------+------+---------------+------------------+------------+ 
1 row in set (0.00 sec)