MySQL计数行

如何根据行的内容计算行数？ 假设我有这样的桌子

[表a]

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 guys

使用子查询并计算结果

在有缺陷的“psuedo sql”中：

当我说有缺陷时，我的意思是我没有尝试过这个，我只是想让大家明白我的想法，而不是给你一些东西来复制粘贴

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，因为我不太明白你想要达到什么目的。

根据Matthew的回答，我建议你使用联合和分组。例如：

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

您需要合并这些结果以获得一个结果集。

类似于：

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

---

。。。这里有一个

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（假设没有比你显示的更多的记录）：

---

您可能需要在select子句.tq中使用

表a.ID\u COMPANY

！它解决了。。事实上，我想得到多少用户帐户为每个注册公司在我的网站。thx=）如果我的答案解决了您的问题，您应该单击答案旁边的复选框将其标记为“已接受”答案。

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;

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