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所以我有一个设计，将人们分组，这些组决定每个人保存哪些属性。

因此，如果稍后我决定要拥有 VIP 客户并记录他们的 VIP 号码，我可以创建一个名为“VIP 成员”的新组，其属性为“VIP_Number”，并将这些客户放在该组中。

因此，员工和其他客户没有与他们无关的字段。

现在，从给定人员的单独表中获取所有必需数据的最佳方法是什么？我需要一个支持以后添加的数据表的解决方案。

这是桌子设计。

笔记：

  • 'd' 前缀用于数据表
  • 'x' 前缀用于交叉引用表
  • 'p' 前缀用于属性
  • 我使用了不同长度的 ID 来帮助提高可读性
  • John Smith 是员工，因此也是人。
  • Anna Backhouse 是一个客户，因此也是一个人。
  • Mr Both 既是员工又是客户，因此也是人。

d人

ID                   pFirst_Name          pLast_Name
----------------------------------------------------------
001                  John                 Smith
002                  Anna                 Backhouse
003                  Mr                   Both

d员工

ID                   pBadge_Code
----------------------------------------------------------
01                   MB2012
02                   JS2012

d客户

ID                   pPhone_Number
----------------------------------------------------------
01                   1800 backhouse
02                   1800 both

d组

ID                   pGroup_Name
----------------------------------------------------------
1                    People
2                    Employees
3                    Customers

x组

dGROUPS_ID_parent    dGROUPS_ID_child
----------------------------------------------------------
1                    1
1                    2
1                    3

x人

dPeople_ID           dGROUPS_ID
----------------------------------------------------------
001                  2
002                  3
003                  2
003                  3

x员工

dPeople_ID           dEmployees_ID
----------------------------------------------------------
001                  02
003                  01

x客户

dPeople_ID           dCustomers_ID
----------------------------------------------------------
002                  01
003                  02

查找 Mr Both 时的期望结果：

ID    pFirst_Name  pLast_Name  pBadge_Code  pPhone_Number
----------------------------------------------------------
003   Mr           Both        MB2012       1800 both

查找 John Smith 时的期望结果：

ID    pFirst_Name  pLast_Name  pBadge_Code  
----------------------------------------------------------
001   John         Smith       JS2012

查找 Anna Backhouse 所需的结果：

ID    pFirst_Name  pLast_Name  pPhone_Number
----------------------------------------------------------
002   Anna         Backhouse   1800 backhouse

我可以获得他们正在使用的所有组的列表：

SELECT `dGROUPS`.`ID` AS `Group ID`, `dGROUPS`.`pGroup_Name` AS `Group Name` FROM `dGROUPS`, `xGROUPS` WHERE `xGROUPS`.`dGROUPS_ID_parent` = `dGROUPS`.`ID` AND (`xGROUPS`.`dGROUPS_ID_child` IN (SELECT `dGROUPS_ID` FROM `xPeople` WHERE dPeople_ID = 003))

当我有更多的子组时，这很有帮助，例如人员 > 员工 > 经理

我想我需要运行这个查询并使用它动态地形成另一个查询以获得所需的结果。我不介意使用两个查询，但是首选一个。

也许有使用视图的解决方案？

我正在使用 PHP 和 MySQL。

4

2 回答 2

0

我修改了 select 子句，星号选择所有列，甚至是后来添加到表中的新列。

两位先生：

  --select dp.ID, dp.pFirst_Name, dp.pLast_Name, de.pBadge_Code, dc.pPhone_Number
    select dp.*, dp.*, dp.*, de.*, dc.*
    from dPeople dp join xCustomers xc
      on dp.id = xc.dPeople_ID
    join dCustomers dc
      on xc.dCustomers_ID = dc.ID
    join xEmployees xe
      on dp.id = xe.dPeople_ID
    join dEmployees de
      on  xe.dEmployees_ID = de.ID
于 2012-07-20T07:52:16.213 回答
0

If you don't mind about having empty columns (you could handle them in the php side) you can get the desired result without using dynamic queries. This would also allow you to throw away grouping tables: dGROUPS and xGROUPS don't give any added value for just this purpose the relation among entities is already given by the actual data, they only generate Update anomalies.

This a sample made with postgres (but you can do the same with mysql).

postgres=# with dPeople(ID, pFirst_Name, pLast_Name) as (
postgres(#              values('001','John','Smith')
postgres(#              union all
postgres(#              values('002','Anna','Backhouse')
postgres(#              union all
postgres(#              values('003','Mr','Both')
postgres(#      ), dEmployees(ID, pBadge_Code) as (
postgres(#              values('01','MB2012')
postgres(#              union all
postgres(#              values('02','JS2012')
postgres(#      ), dCustomers (ID, pPhone_Number) as (
postgres(#              values('01','1800 backhouse')
postgres(#              union all
postgres(#              values('02','1800 both')
postgres(#      ), xPeople (dPeople_ID, dGROUPS_ID) as (
postgres(#              values('001','2')
postgres(#              union all
postgres(#              values('002','3')
postgres(#              union all
postgres(#              values('003','2')
postgres(#              union all
postgres(#              values('003','3')
postgres(#      ), xEmployees (dPeople_ID, dEmployees_ID) as (
postgres(#              values('001','02')
postgres(#              union all
postgres(#              values('003','01')
postgres(#      ), xCustomers (dPeople_ID, dCustomers_ID) as (
postgres(#              values('002','01')
postgres(#              union all
postgres(#              values('003','02')
postgres(#      )
postgres-# select *
postgres-# from (
postgres(#              select ID as dPeople_ID, pFirst_Name, pLast_Name
postgres(#              from dPeople
postgres(#      ) dPeople
postgres-#      left outer join (
postgres(#                      select dPeople_ID, dEmployees_ID, pBadge_Code
postgres(#                      from xEmployees
postgres(#                              join (
postgres(#                                              select ID as dEmployees_ID, pBadge_Code
postgres(#                                              from dEmployees
postgres(#                                      ) as dEmployees using (dEmployees_ID)
postgres(#              ) as Employees using (dPeople_ID)
postgres-#      left outer join (
postgres(#                      select dPeople_ID, dCustomers_ID,  pPhone_Number
postgres(#                      from xCustomers
postgres(#                              join (
postgres(#                                              select ID as dCustomers_ID, pPhone_Number
postgres(#                                              from dCustomers
postgres(#                                      ) as Customers using (dCustomers_ID)
postgres(#              ) as Customers using (dPeople_ID)
postgres-# ;
 dpeople_id | pfirst_name | plast_name | demployees_id | pbadge_code | dcustomers_id | pphone_number
------------+-------------+------------+---------------+-------------+---------------+----------------
 001        | John        | Smith      | 02            | JS2012      |               |
 002        | Anna        | Backhouse  |               |             | 01            | 1800 backhouse
 003        | Mr          | Both       | 01            | MB2012      | 02            | 1800 both
(3 rows)
于 2012-07-20T12:16:47.817 回答