Mysql 计算多个值的不同行数 我们考虑一下这张表，说明一个人买了多少次房产。< /P> +--------+----------+ | user | property | +--------+----------+ | john | car | | john | car | | john | house | | peter | car | | peter | car | | amanda | house | | amanda | house | +--------+----------+

我需要知道一辆车买了多少次，一栋房子买了多少次，等等，像这样的：

+----------+---+---+
| property | 1 | 2 |
+----------+---+---+
| cars     | 4 | 2 |
| house    | 3 | 1 |
+----------+---+---+

• 一辆车买了多少次四个，两个给彼得，两个给约翰
• 一辆车买了两次有多少次两个，为同一个人
• 一栋房子买了多少次三个，两个给阿曼达，一个给约翰
• 一栋房子买了两次有多少次？仅一次，对于amanda

只有使用SQL查询才能做到这一点吗？

• 我不在乎性能或老套的方式
• 有两个以上的频率
• 一个人可以购买房产的时间是固定的（5），因此在查询中手动指定列是没有问题的。我的意思是这样做没有问题：

  SELECT /* ... */ AS 1, /* ... */ AS 2, /* ... */, AS 3 /* ... */
  

试试这个

    SELECT property , count(property) as bought_total , count(distinct(user)) bought_per_user
    FROM Table1
    GROUP BY property

输出将是这样的

  PROPERTY  |   BOUGHT_TOTAL     |  BOUGHT_PER_USER
  ________________________________________________________
    car     |            4       |       2
    house   |            3       |       2

---

您应该能够使用子选择来完成此操作

SELECT property, user, COUNT(*) FROM purchases GROUP BY property, user;

将返回所需的完整分组数据集。然后，您需要查看不同的频率：

SELECT property, freq, COUNT(*) FROM (SELECT property, user, COUNT(*) freq FROM purchases GROUP BY property, user) AS foo GROUP BY property, freq;

它的格式与您演示的不完全相同，但它返回数据

SELECT DISTINCT @pr := prop,
    (SELECT COUNT(1) FROM tbl WHERE prop = @pr LIMIT 1),
    (SELECT COUNT(1) FROM 
        (SELECT *, COUNT(*) cnt
        FROM tbl
        GROUP BY usr, prop
        HAVING cnt = 2) as tmp
        WHERE `tmp`.prop = @pr LIMIT 1)
FROM tbl;

是的，这不是最好的方法；但是，嘿，你得到了想要的答案

此外，它还将生成表中任何类型属性的结果

小提琴链接

---

附言：60次尝试O_O

自从你发布问题后，我就在这里。好的…
这里有一种方法可以完全按照您的要求进行，只需分组和计数。
诀窍是我将用户列和属性列连接起来，为每一列生成一个唯一的“id”，如果我们可以这样称呼它的话。它应该独立于购买计数工作

SELECT C.`property`, COUNT(C.`property`), D.`pcount` from `purchases` C
LEFT JOIN(
  SELECT A.`property`, B.`pcount` FROM `purchases` A
  LEFT JOIN (
    SELECT `property`,
           CONCAT(`user`, `property`) as conc,
           COUNT(CONCAT(`user`, `property`)) as pcount
    FROM `purchases` GROUP BY CONCAT(`user`, `property`)
  ) B
  ON A.`property` = B.`property`
  GROUP BY B.pcount
) D
ON C.`property` = D.`property`
GROUP BY C.`property`

MySQL 5.5.30架构设置：

CREATE TABLE Table1
    (`user` varchar(6), `property` varchar(5))
;

INSERT INTO Table1
    (`user`, `property`)
VALUES
    ('john', 'car'),
    ('john', 'car'),
    ('john', 'house'),
    ('peter', 'car'),
    ('peter', 'car'),
    ('amanda', 'house'),
    ('amanda', 'house')
;

select t.property, t.total, c1.cnt as c1, c2.cnt as c2, c3.cnt as c3
from
  (select 
    t.property , 
    count(t.property) as total
  from Table1 t
  group by t.property
  ) as t
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 1
        ) as i1
      group by property
  ) as c1 on t.property = c1.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 2
        ) as i2
      group by property
  ) as c2 on t.property = c2.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 3
        ) as i3
      group by property
  ) as c3 on t.property = c3.property

| PROPERTY | TOTAL |     C1 | C2 |     C3 |
-------------------------------------------
|      car |     4 | (null) |  2 | (null) |
|    house |     3 |      1 |  1 | (null) |

查询1：

CREATE TABLE Table1
    (`user` varchar(6), `property` varchar(5))
;

INSERT INTO Table1
    (`user`, `property`)
VALUES
    ('john', 'car'),
    ('john', 'car'),
    ('john', 'house'),
    ('peter', 'car'),
    ('peter', 'car'),
    ('amanda', 'house'),
    ('amanda', 'house')
;

select t.property, t.total, c1.cnt as c1, c2.cnt as c2, c3.cnt as c3
from
  (select 
    t.property , 
    count(t.property) as total
  from Table1 t
  group by t.property
  ) as t
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 1
        ) as i1
      group by property
  ) as c1 on t.property = c1.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 2
        ) as i2
      group by property
  ) as c2 on t.property = c2.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 3
        ) as i3
      group by property
  ) as c3 on t.property = c3.property

| PROPERTY | TOTAL |     C1 | C2 |     C3 |
-------------------------------------------
|      car |     4 | (null) |  2 | (null) |
|    house |     3 |      1 |  1 | (null) |

：

CREATE TABLE Table1
    (`user` varchar(6), `property` varchar(5))
;

INSERT INTO Table1
    (`user`, `property`)
VALUES
    ('john', 'car'),
    ('john', 'car'),
    ('john', 'house'),
    ('peter', 'car'),
    ('peter', 'car'),
    ('amanda', 'house'),
    ('amanda', 'house')
;

select t.property, t.total, c1.cnt as c1, c2.cnt as c2, c3.cnt as c3
from
  (select 
    t.property , 
    count(t.property) as total
  from Table1 t
  group by t.property
  ) as t
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 1
        ) as i1
      group by property
  ) as c1 on t.property = c1.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 2
        ) as i2
      group by property
  ) as c2 on t.property = c2.property
  left join (
      select property, count(*) as cnt
      from (
          select 
            property, user, count(*) as cnt
          from table1
          group by property, user
          having count(*) = 3
        ) as i3
      group by property
  ) as c3 on t.property = c3.property

| PROPERTY | TOTAL |     C1 | C2 |     C3 |
-------------------------------------------
|      car |     4 | (null) |  2 | (null) |
|    house |     3 |      1 |  1 | (null) |

你可以试试下面的

SELECT COUNT(TABLE1.PROPERTY) AS COUNT, PROPERTY.USER FROM TABLE1
INNER JOIN (SELECT DISTINCT PROPERTY, USER FROM TABLE1) AS PROPERTY
ON PROPERTY.PROPERTY = TABLE1.PROPERTY
AND PROPERTY.USER = TABLE1.USER
GROUP BY TABLE1.USER, PROPERTY.PROPERTRY

---

在MySQL中进行了类似的测试

我希望这能帮助您…..让我们先创建一个表：

创建表属性（用户varchar（最大值），属性varchar（最大值））

插入道具值（'john'，'car'），插入道具值（'john'，'car'）， 插入道具值（'john'，'house'），插入道具值（'peter'，'car'）

插入道具值（'peter'，'car'），插入道具值（'amanda'，'house'）， 插入道具值（'amanda'、'house'）

1） 这辆车买了多少次

回答：从属性为'car'的道具中选择计数（属性） (四)

2） 一辆车买了两次有多少次

ANS：选择用户，从属性=‘汽车’组中按用户计数（属性） 具有计数（属性）=2

2-john 2-彼得

3） 一栋房子买了多少次

ANS：从prop中选择COUNT（property），其中property='house' (3) 4） 一栋房子买了两次有多少次？ ANS：按用户从prop where property='house'组中选择用户、计数（属性） 计数（属性）<=2 2-阿曼达

---

1-john

谢谢您抽出时间。这给了我不同用户购买房产的次数，但这不是我想要的。请注意，结果表与示例输出表不同。这太棒了！不过需要注意的是，它告诉我某个用户购买房产的次数。例如，如果amanda买了另一栋房子，那么一栋房子被买两次的次数将为0（即使amanda之前买了两次房子）。尽管如此，距离足够近，它仍然非常有用。如果没有其他弹出窗口，这就是正确答案。您可以将

具有cnt=2

更改为

具有cnt>=2

。或增加另一条款；其中设置cnt=3将给出所有三次购买的产品：P