好歹選擇行列 - SQL服務器

我想知道是否存在某種方式來選擇表波紋管......好歹選擇行列 - SQL服務器

ID | DOCNo | Date 

1 | 20 | 12-DEC-12 
2 | 21 | 02-NOV-12 
3 | 20 | 18-JAN-13 
4 | 20 | 10-APR-13 
5 | 22 | 12-DEC-12

做這樣的事情在SQL Server 2008：

DOCNo | ID1 | Date1  | ID2 | Date2  | ID3 | Date3 

20 | 1 | 12-DEC-12 | 3 | 18-JAN-13 | 4 | 10-APR-13 
21 | 2 | 02-NOV-12 | NULL | NULL | NULL | NULL  
22 | 5 | 12-DEC-12 | NULL | NULL | NULL | NULL

*的DOCNo將永遠不會有超過3個記錄表

我使用SQL Server 2008的

謝謝!!!

來源

2013-10-22 Vitor Moscolini

是'SELECT'表中有一定數量的IDn。含義ID1，ID2，ID3是它停止或無限的地方。你可以有7,10,15 ... – logixologist

其實我剛剛看到：'DOCNo將永遠不會有超過3個記錄在表中.... ....它必須是一個查詢或者它可以是一個存儲過程。 ...我可以在使用臨時表的過程中想到一種方法。 – logixologist

嗨logixologist，感謝您的回覆......您認爲可以成爲一個查詢嗎？這對我來說是最好的選擇... –

SQL Fiddle

MS SQL Server 2008的架構設置：

CREATE TABLE Table1 
    ([ID] int, [DOCNo] int, [Date] varchar(9)) 
; 

INSERT INTO Table1 
    ([ID], [DOCNo], [Date]) 
VALUES 
    (1, 20, '12-DEC-12'), 
    (2, 21, '02-NOV-12'), 
    (3, 20, '18-JAN-13'), 
    (4, 20, '10-APR-13'), 
    (5, 22, '12-DEC-12'), 
    (6, 23, '21-JAN-13'), 
    (7, 23, '24-FEB-13') 
;

查詢1：

SELECT r1.DOCNo, 
     r1.ID as ID1, r1.Date as Date1, 
     r2.ID as ID2, r2.Date as Date2, 
     r3.ID as ID3, r3.Date as Date3 
FROM Table1 r1 
LEFT OUTER JOIN Table1 r2 ON r1.DOCNo = r2.DOCNo AND r1.ID < r2.ID 
LEFT OUTER JOIN Table1 r3 ON r1.DOCNo = r3.DOCNo AND r1.ID < r3.ID 
               AND r2.ID < r3.ID 
WHERE r1.ID = (SELECT min(ID) 
       FROM Table1 
       WHERE DOCNo = r1.DOCNo 
       GROUP BY DOCNo) 
AND (CASE WHEN (SELECT count(*) 
        FROM Table1 
       WHERE DOCNo = r1.DOCNo 
       GROUP BY DOCNo) = 2 THEN r2.ID 
      WHEN (SELECT count(*) 
        FROM Table1 
       WHERE DOCNo = r1.DOCNo 
       GROUP BY DOCNo) = 3 THEN r3.ID ELSE 1 END) IS NOT NULL

Results：

| DOCNO | ID1 |  DATE1 | ID2 |  DATE2 | ID3 |  DATE3 | 
|-------|-----|-----------|--------|-----------|--------|-----------| 
| 20 | 1 | 12-DEC-12 |  3 | 18-JAN-13 |  4 | 10-APR-13 | 
| 21 | 2 | 02-NOV-12 | (null) | (null) | (null) | (null) | 
| 22 | 5 | 12-DEC-12 | (null) | (null) | (null) | (null) | 
| 23 | 6 | 21-JAN-13 |  7 | 24-FEB-13 | (null) | (null) |

來源

2013-10-23 01:02:10

一個簡單的方法來獲得最終的結果是使用row_number()生成每個docno序列，然後用一個CASE表達式應用聚合函數：

select docno, 
    max(case when seq = 1 then id end) id1, 
    max(case when seq = 1 then date end) date1, 
    max(case when seq = 2 then id end) id2, 
    max(case when seq = 2 then date end) date2, 
    max(case when seq = 3 then id end) id3, 
    max(case when seq = 3 then date end) date3 
from 
(
    select id, docno, date, 
    row_number() over(partition by docno order by date) seq 
    from yourtable 
) d 
group by docno;

見SQL Fiddle with Demo

來源

2013-10-23 01:04:02 Taryn

好歹選擇行列 - SQL服務器

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