Question

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＆amp;和新的列顺序合并项目。按事务划分如下。 [项目将通过订单ASC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，该怎么办。

    +-----------------+--------+---------+------------------+
    |     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       |
    +-----------------+--------+---------+------------------+

Answer 1

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

与[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

Answer 2

试试这种方式

   final Class[] args = new Class[1];
   args[0] = Context.class;
   final Method doSomething = classToLoad.getMethod("doSomething",    args);
   final Object myInstance = classToLoad.newInstance();
   doSomething.invoke(myInstance, this);

Sql将行数据连接到列中

2 个答案: