Question

我最近一直在刷我的MySQL，我需要创建一个包含分层数据的数据库。

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，可以雇用或受雇于其他人。这些人中的每一个都可能有设备检查，每件设备必须有名称，描述和更换部件清单，每个更换部件必须有成本等等。

我看到关闭表的大多数示例都集中在处理论坛或线程注释的方式上。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 |

他们可以查询如下：

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 |

具有多种数据类型的闭包表？

1 个答案: