如何複製數據庫記錄並跟蹤其新ID

Question

我有三個表：年，員工，職位。假設我已經在這些表中有這些數據。

years: 
---------------- 
| id | name | 
---------------- 
| 1 | 2011 | 
---------------- 

positions: 
------------------------------ 
| id | name  | year_id | 
------------------------------ 
| 1 | Director | 1  | 
| 2 | Manager  | 1  | 
------------------------------ 

employees: 
--------------------------------------------------------- 
| id | name      | position_id | year_id | 
--------------------------------------------------------- 
| 1 | Employee A (Director) | 1   | 1  | 
| 2 | Employee B (Manager)  | 2   | 1  | 
--------------------------------------------------------- 

========================================

年表是一箇中心點。 如果我插入新的年份記錄，我還必須複製與上一年相關的所有職位和員工。

所以，如果我插入2012年入裏表，數據被假設是這樣的：

years: 
---------------- 
| id | name | 
---------------- 
| 1 | 2011 | 
| 2 | 2012 | 
---------------- 

positions: 
------------------------------ 
| id | name  | year_id | 
------------------------------ 
| 1 | Director | 1  | 
| 2 | Manager  | 1  | 
| 3 | Director | 2  | 
| 4 | Manager  | 2  | 
------------------------------ 

employees: 
--------------------------------------------------------- 
| id | name      | position_id | year_id | 
--------------------------------------------------------- 
| 1 | Employee A (Director) | 1   | 1  | 
| 2 | Employee B (Manager)  | 2   | 1  | 
| 3 | Employee A (Director) | 3 (?) | 2  | 
| 4 | Employee B (Manager)  | 4 (?) | 2  | 
--------------------------------------------------------- 

注意employees表的第三和第四行中的問號。

我使用這些查詢在新的一年中插入和複製所有相關的職位和員工：

// Insert new year record 
INSERT INTO years (name) VALUES (2012); 

// Get last inserted year ID 
$inserted_year_id = .......... // skipped 

// Copy positions 
INSERT INTO positions (name, year_id) SELECT name, $inserted_year_id AS last_year_id FROM positions WHERE year_id = 1; 

// Copy employees 
INSERT INTO employees (name, position_id, year_id) SELECT name, position_id, $inserted_year_id AS last_year_id FROM employees WHERE year_id = 1; 

的問題是在複製的員工查詢。我無法找到獲取或跟蹤職位新ID的方法。

有沒有一個簡單的方法來做到這一點？

非常感謝。

Answer 1

你的數據模型存在嚴重缺陷，可能需要一個完整的檢修，但如果你堅持要複製的數據像你描述的，這應該做的伎倆：

// Copy employees 
INSERT INTO employees (name, position_id, year_id) 
SELECT name, new_positions.id, $inserted_year_id AS last_year_id 
FROM employees 
JOIN positions AS old_positions ON old_positions.id = employees.position_id 
           AND old_positions.year_id = employees.year_id 
JOIN positions AS new_positions ON new_positions.name = old_positions.name 
           AND new_positions.year_id = $inserted_year_id 
WHERE employees.year_id = 1 

Answer 2

我想你應該閱讀有關database normalization。複製數據導致維護問題和錯誤的報告。

如果您採用了與以下不同的設計，那麼在員工更改職位，終止職位或停止職位之前，不會插入任何內容。還有很多其他方法可以解決這個問題，但您應該儘量減少冗餘（即每個員工只有一份副本），然後分別跟蹤隨時間變化的數據。在嘗試實現像這樣的事情之前，還請閱讀foreign keys。

positions: 
-- If you are keeping track of the years that each position is active, 
-- using dates provides simplicity. Note: this design assumes that positions 
-- are never reactivated after being deactivated. 
------------------------------------------------ 
| id | name  | DateActive | DateInactive | 
------------------------------------------------ 
| 1 | Director | 01/01/2011 |    | 
| 2 | Manager  | 01/01/2011 |    | 
------------------------------------------------ 

employees: 
--------------------------------------------------------------- 
| id | name      | DateHired | DateTerminated | 
--------------------------------------------------------------- 
| 1 | Employee A    | 01/01/2011 |    | 
| 2 | Employee B    | 01/01/2011 |    | 
| 3 | Employee C    | 01/01/2011 | 10/01/2012  | 
--------------------------------------------------------------- 

EmployeePositionRelationships 
--If you are keeping track of time that each employee held a position 
-- Employee A has been a Director since 1/1/2011 
-- Employee B was a Manager from 1/1/2011 to 10/6/2012. Then they became a Director 
-- Employee B was a Manager from 1/1/2011 to 10/1/2012. Then they were terminated 
-------------------------------------------------------- 
EmployeeId | PositionId | DateStarted | DateEnded  | 
-------------------------------------------------------- 
1   | 1   | 01/01/2011 |    | 
2   | 2   | 01/01/2011 | 10/6/2012  | 
3   | 2   | 01/01/2011 | 10/1/2012  | 
2   | 1   | 10/6/2012 |    | 
--------------------------------------------------------