Question

简单而快速的问题，我有这些表：

//table people
| pe_id | pe_name |
| 1  | Foo  |
| 2  | Bar  |
//orders table
| ord_id | pe_id | ord_title   |
|   1    |   1   | First order |
|   2    |   2   | Order two   |
|   3    |   2   | Third order |
//items table
| item_id | ord_id | pe_id | title  |
|   1     |   1    |   1   | Apple  |
|   2     |   1    |   1   | Pear   |
|   3     |   2    |   2   | Apple  |
|   4     |   3    |   2   | Orange |
|   5     |   3    |   2   | Coke   |
|   6     |   3    |   2   | Cake   |

我需要查询列出所有人，计算订单数量和总项目数量，例如：

| pe_name | num_orders | num_items |
| Foo  |    1       |   2       |
| Bar  |    2       |   4       |

但我不能让它发挥作用！我试过了

SELECT
    people.pe_name,
    COUNT(orders.ord_id) AS num_orders,
    COUNT(items.item_id) AS num_items
FROM
    people
    INNER JOIN orders ON (orders.pe_id = people.pe_id)
    INNER JOIN items ON items.pe_id = people.pe_id
GROUP BY
    people.pe_id;

但这会使num_*值返回错误：

| name | num_orders | num_items |
| Foo  |    2       |   2       |
| Bar  |    8       |   8       |

我注意到，如果我尝试加入一张桌子，那就可以了：

SELECT
    people.pe_name,
    COUNT(orders.ord_id) AS num_orders
FROM
    people
    INNER JOIN orders ON (orders.pe_id = people.pe_id)
GROUP BY
    people.pe_id;

//give me:
| pe_name | num_orders |
| Foo     |          1 |
| Bar     |          2 |

//and:
SELECT
    people.pe_name,
    COUNT(items.item_id) AS num_items
FROM
    people
    INNER JOIN items ON (items.pe_id = people.pe_id)
GROUP BY
    people.pe_id;
//output:
| pe_name | num_items |
| Foo     |         2 |
| Bar     |         4 |

如何将这两个查询合并为一个？

Answer 1

将项目与订单联系起来比使用人员更有意义！

SELECT
    people.pe_name,
    COUNT(distinct orders.ord_id) AS num_orders,
    COUNT(items.item_id) AS num_items
FROM
    people
    INNER JOIN orders ON orders.pe_id = people.pe_id
         INNER JOIN items ON items.ord_id = orders.ord_id
GROUP BY
    people.pe_id;

与人民一起加入物品会引起很多疑惑。例如，3号蛋糕物品将通过人与人之间的联系与订单2相关联，您不希望这种情况发生!!

所以：

1-您需要很好地理解您的架构。物品是订单的链接，而不是人。

2-您需要为一个人计算不同的订单，否则您将计算与订单一样多的商品。

Answer 2

正如弗兰克指出的那样，你需要使用DISTINCT。此外，由于您使用的是复合主键（完全没问题，BTW），因此您需要确保在连接中使用整个键：

SELECT
    P.pe_name,
    COUNT(DISTINCT O.ord_id) AS num_orders,
    COUNT(I.item_id) AS num_items
FROM
    People P
INNER JOIN Orders O ON
    O.pe_id = P.pe_id
INNER JOIN Items I ON
    I.ord_id = O.ord_id AND
    I.pe_id = O.pe_id
GROUP BY
    P.pe_name

如果没有I.ord_id = O.ord_id，则会将每个项目行连接到一个人的每个订单行。

Answer 3

我试着把两者分开， count（distinct ord.ord_id）为num_order，将（distinct items.item_id）计为num items

它的工作：）

    SELECT
         people.pe_name,
         COUNT(distinct orders.ord_id) AS num_orders,
         COUNT(distinct items.item_id) AS num_items
    FROM
         people
         INNER JOIN orders ON (orders.pe_id = people.pe_id)
         INNER JOIN items ON items.pe_id = people.pe_id
    GROUP BY
         people.pe_id;

感谢线程帮助：）

Answer 4

您的解决方案几乎是正确的。你可以添加DISTINCT：

SELECT
    people.pe_name,
    COUNT(distinct orders.ord_id) AS num_orders,
    COUNT(items.item_id) AS num_items
FROM
    people
    INNER JOIN orders ON (orders.pe_id = people.pe_id)
    INNER JOIN items ON items.pe_id = people.pe_id
GROUP BY
    people.pe_id;

Answer 5

需要了解JOIN或一系列JOIN对一组数据的作用。使用strae的帖子，pe_id为1加上相应的订单和pe_id = 1上的项目将为您提供以下数据来“选择”：

[表人部分] [表订单部分] [表项部分]

  | people.pe_id | people.pe_name | orders.ord_id | orders.pe_id | orders.ord_title | item.item_id | item.ord_id | item.pe_id | item.title |

  | 1 | Foo | 1 | 1 |第一顺序| 1 | 1 | 1 | Apple |
  | 1 | Foo | 1 | 1 |第一顺序| 2 | 1 | 1 |梨|

联接基本上是所有表格的笛卡尔积。您基本上可以从中选择数据集，这就是您需要对orders.ord_id和items.item_id进行独特计数的原因。否则两个计数都将产生2 - 因为你实际上有2行可供选择。

在三个表上使用count（）进行内连接

5 个答案: