![](/img/trans.png)
[英]SQL : Concatenate data from a column depending on another column and reading row above
[英]Sql concatenate of row data into column
数据集:
+-----------------+--------+---------+
| TransNo | Serial | Project |
+-----------------+--------+---------+
| A00000000000001 | 1 | 100 |
| A00000000000001 | 2 | 200 |
| A00000000000001 | 3 | 201 |
| A00000000000001 | 4 | 101 |
| A00000000000002 | 1 | 100 |
| A00000000000002 | 2 | 101 |
| A00000000000003 | 1 | 100 |
| A00000000000003 | 2 | 200 |
| A00000000000004 | 1 | 200 |
| A00000000000004 | 2 | 100 |
| A00000000000005 | 1 | 101 |
| A00000000000005 | 2 | 100 |
+-----------------+--------+---------+
我希望按项目和按事务分区按新的列顺序合并项目,如下所示。 [项目将通过ASC订单合并]
输出:
+-----------------+--------+---------+------------------+
| TransNo | Serial | Project | CProject |
+-----------------+--------+---------+------------------+
| A00000000000001 | 1 | 100 | 100101200201 |
| A00000000000001 | 2 | 200 | 100101200201 |
| A00000000000001 | 3 | 201 | 100101200201 |
| A00000000000001 | 4 | 101 | 100101200201 |
| A00000000000002 | 1 | 100 | 100101 |
| A00000000000002 | 2 | 101 | 100101 |
| A00000000000005 | 1 | 101 | 100101 |
| A00000000000005 | 2 | 100 | 100101 |
| A00000000000003 | 1 | 100 | 100200 |
| A00000000000003 | 2 | 200 | 100200 |
| A00000000000004 | 1 | 200 | 100200 |
| A00000000000004 | 2 | 100 | 100200 |
+-----------------+--------+---------+------------------+
更新1:
如果我想按Serial而不是项目输出订单怎么办。
+-----------------+--------+---------+------------------+
| TransNo | Serial | Project | CProject |
+-----------------+--------+---------+------------------+
| A00000000000001 | 1 | 100 | 100200201101|
| A00000000000001 | 2 | 200 | 100200201101|
| A00000000000001 | 3 | 201 | 100200201101|
| A00000000000001 | 4 | 101 | 100200201101|
| A00000000000002 | 1 | 100 | 100101 |
| A00000000000002 | 2 | 101 | 100101 |
| A00000000000005 | 1 | 101 | 101100 |
| A00000000000005 | 2 | 100 | 101100 |
| A00000000000003 | 1 | 100 | 100200 |
| A00000000000003 | 2 | 200 | 100200 |
| A00000000000004 | 1 | 200 | 200100 |
| A00000000000004 | 2 | 100 | 200100 |
+-----------------+--------+---------+------------------+
DECLARE @t TABLE (
TransNo VARCHAR(20),
Serial INT,
Project INT
)
INSERT INTO @t (TransNo, Serial, Project)
VALUES
('A00000000000001', 1, 100),
('A00000000000001', 2, 200),
('A00000000000001', 3, 201),
('A00000000000001', 4, 101),
('A00000000000002', 1, 100),
('A00000000000002', 2, 101),
('A00000000000003', 1, 100),
('A00000000000003', 2, 200),
('A00000000000004', 1, 200),
('A00000000000004', 2, 100),
('A00000000000005', 1, 101),
('A00000000000005', 2, 100)
SELECT *, CProject = (
SELECT DISTINCT [text()] = t2.Project
FROM @t t2
WHERE t2.TransNo = t1.TransNo
ORDER BY t2.Project
FOR XML PATH('')
)
FROM @t t1
输出 -
TransNo Serial Project CProject
-------------------- ----------- ----------- --------------
A00000000000001 1 100 100101200201
A00000000000001 2 200 100101200201
A00000000000001 3 201 100101200201
A00000000000001 4 101 100101200201
A00000000000002 1 100 100101
A00000000000002 2 101 100101
A00000000000003 1 100 100200
A00000000000003 2 200 100200
A00000000000004 1 200 100200
A00000000000004 2 100 100200
A00000000000005 1 101 100101
A00000000000005 2 100 100101
[text()]
-
100101200201
没有[text()]
-
<Project>100</Project><Project>101</Project><Project>200</Project><Project>201</Project>
更多详情 - http://www.codeproject.com/Articles/691102/String-Aggregation-in-the-World-of-SQL-Server
更新 -
SELECT *, CProject = (
SELECT [text()] = t2.Project
FROM (
SELECT t2.Project, Serial = MIN(t2.Serial)
FROM @t t2
WHERE t2.TransNo = t1.TransNo
GROUP BY t2.Project
) t2
ORDER BY t2.Serial
FOR XML PATH('')
)
FROM @t t1
试试这种方式
SELECT t1.*,t2.CProject
FROM t AS t1
JOIN (
SELECT SS.TransNo, (
SELECT ' ' + US.Project
FROM t US
WHERE US.TransNo = SS.TransNo
FOR XML PATH('')
) CProject
FROM t SS
GROUP BY SS.TransNo
ORDER BY SS.Serial
) t2 ON t1.TransNo = t2.TransNo
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