Question

让我们考虑一下这个表格，指明一个人购买房产的次数。

+--------+----------+
|  user  | property |
+--------+----------+
| john   | car      |
| john   | car      |
| john   | house    |
| peter  | car      |
| peter  | car      |
| amanda | house    |
| amanda | house    |
+--------+----------+

我需要知道一次购买汽车的次数，一次购买房屋的次数，等等。这样的事情：

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

购买汽车多少次？四，两个用于 peter ，两个用于 john 。
两次购买汽车多少次？两个，对于同样的人。
买房子多少次？三个，两个用于 amanda ，一个用于 john 。
两次买房子的次数是多少次？对于 amanda

一次

仅使用SQL查询才能实现此目的吗？

我不关心表演或黑客行为。
频率超过两个。
一个人可以购买房产的固定时间（5），因此在查询中手动指定列并不成问题。我的意思是做出类似的事情没有问题：
```
SELECT /* ... */ AS 1, /* ... */ AS 2, /* ... */, AS 3 /* ... */
```

Answer 1

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，你可以得到你想要的答案。

此外，它还会为您表格中的任何类型的属性生成结果。

小提琴链接lies here。

P.S。：60次尝试O_O

Answer 2

我在这里发布了这个问题。好的... ... 这是一种完全按照你的要求去做的方法，只有组和数量诀窍是我连接用户和属性列，为每个列生成一个唯一的“id”，如果我们可以调用它。它应该独立于购买计数。

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`

Answer 3

SQL Fiddle

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')
;

查询1 ：

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

<强> Results ：

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

Answer 4

您可以尝试跟随。

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中测试类似

Answer 5

试试这个

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

输出就像那样

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

DEMO SQL FIDDLE HERE

Answer 6

你应该能够通过子选择来做到这一点。

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

将返回您想要的完整分组数据集。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;

它并不完全符合您所说明的格式，但会返回数据

Answer 7

我希望这可以帮助你......让我们先创建一个表：

create table prop（user varchar（max），property varchar（max））

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）买了多少次汽车？

ANS：从道具中选择count（属性），其中property ='car' （4）

2）两次购买汽车的次数是多少次？

ANS：从prop中选择用户，COUNT（属性），其中property ='car'按用户分组有COUNT（财产）= 2

2 - 约翰 2-彼得

3）房子买了多少次？

ANS：从prop中选择COUNT（属性），其中property ='house' （3） 4）两次买房子的次数是多少次？ ANS：从prop中选择用户，COUNT（属性），其中property ='house'按用户分组有COUNT（财产）＆lt; = 2 2-阿曼达 1-约翰

计算多个值的不同行数

7 个答案: