具有多种数据类型的闭包表

Question

我最近一直在复习我的MySQL，我需要用分层数据创建一个数据库。

我有几种不同类型的数据需要以树格式表示，但不知道如何去做。

例如，假设我有一个人，他可以雇用或被其他人雇用。这些人中的每一个都可以向他们检查设备，每件设备必须有名称、描述和更换零件清单，每个更换零件必须有成本等。

我看到的大多数闭包表示例都集中在它们处理论坛或线程评论方面有多棒。如何制作具有多种数据类型的闭包表？

Answer 1

这是一个快速而肮脏的例子：

select * from person
| pID | name      | employedBy |
+-----+-----------+------------+
|   1 | John Doe  |          2 |
|   2 | Joe Smith |       NULL |
|   3 | Meg Ryan  |          3 |
select * from equipment
| eqID | eqName   | eqDescription     | eqOwner | eqCheckedOutTo |
+------+----------+-------------------+---------+----------------+
|    1 | stuff    | just some stuff   |       3 |           NULL |
|    2 | table    | a table           |       1 |           NULL |
|    3 | computer | PC computer       |       3 |              2 |
|    4 | 3table   | table with 3 legs |       2 |           NULL |
select * from parts;
| partID | partName     | partCost |
+--------+--------------+----------+
|      1 | desktop1     |   499.99 |
|      2 | monitor13x13 |   109.95 |
|      3 | windows95    |    10.00 |
|      4 | speakers     |    30.00 |
|      5 | tabletop     |   189.99 |
|      6 | table leg    |    59.99 |
select * from equipmentParts
| epID | eqID | partID | quantity |
+------+------+--------+----------+
|    1 |    3 |      1 |        1 |
|    2 |    3 |      2 |        2 |
|    3 |    3 |      3 |        1 |
|    4 |    2 |      5 |        1 |
|    5 |    2 |      6 |        4 |
|    6 |    4 |      5 |        1 |
|    7 |    4 |      6 |        3 |

您可以像这样查询它们：

select name,eqName,e.eqID,partName,partCost,quantity,(quantity*partCost) AS totCost
from person p
inner join equipment e ON e.eqOwner=p.pID
inner join equipmentParts ep ON ep.eqID=e.eqID
inner join parts pa ON ep.partID=pa.partID
| name      | eqName   | eqID | partName     | partCost | quantity | totCost |
+-----------+----------+------+--------------+----------+----------+---------+
| John Doe  | table    |    2 | tabletop     |   189.99 |        1 |  189.99 |
| John Doe  | table    |    2 | table leg    |    59.99 |        4 |  239.96 |
| Meg Ryan  | computer |    3 | desktop1     |   499.99 |        1 |  499.99 |
| Meg Ryan  | computer |    3 | monitor13x13 |   109.95 |        2 |  219.90 |
| Meg Ryan  | computer |    3 | windows95    |    10.00 |        1 |   10.00 |
| Joe Smith | 3table   |    4 | tabletop     |   189.99 |        1 |  189.99 |
| Joe Smith | 3table   |    4 | table leg    |    59.99 |        3 |  179.97 |

或总结每台设备的成本：

select name,eqName,sum(quantity*partCost) AS totCost
from person p
inner join equipment e ON e.eqOwner=p.pID
inner join equipmentParts ep ON ep.eqID=e.eqID
inner join parts pa ON ep.partID=pa.partID
group by e.eqID
| name      | eqName   | totCost |
+-----------+----------+---------+
| John Doe  | table    |  429.95 |
| Meg Ryan  | computer |  729.89 |
| Joe Smith | 3table   |  369.96 |