Question

以下是我在表格中名为TEMP2的列

[ID],[Week1], [Week2],[Week3],[Week4],[Week5], 
[Week6],[Week7],[Week8],[Week9], [Week10],
[Week11],[Week12],[Week13], [Week14],[Week15],
[Week16],[Week17], [Week18],[Week19],[Week20],
[Week21], [Week22],[Week23],[Week24],[Week25], [Week26]

我想找到每行的前6条记录并显示。

以下是我尝试使用的查询，但未获取预期结果。我在这里做错了什么？

SELECT 
    id, [1], [2], [3],[4], [5], [6]
    FROM 
    (SELECT 
        id, Val, rn
        FROM 
        (SELECT id, Val, Col,
            ROW_NUMBER()OVER (PARTITION BY id ORDER BY Val) 
            AS rn
                FROM 
                TEMP2
    UNPIVOT 
    (Val FOR Col IN ([Week1],
    [Week2],[Week3],[Week4],[Week5],[Week6],[Week7],[Week8],[Week9],
    [Week10],[Week11],[Week12],[Week13],[Week14],[Week15],[Week16],[Week17],
    [Week18],[Week19],[Week20],[Week21],[Week22],[Week23],[Week24],[Week25],
    [Week26]))AS unpvt) AS t
        WHERE 
        t.rn <= 6) AS src
            PIVOT 
            (MAX(Val) FOR rn IN ([1], [2], [3],[4], [5], [6])) AS pvt

我正在使用SQL Server 2008。

Answer 1

正如https://github.com/jcodec/jcodec/blob/master/src/main/java/org/jcodec/api/SequenceEncoder.java在上述评论中所建议的那样，您的PARTITION BY语句中只需要SELECT id, [1], [2], [3], [4], [5], [6] FROM (SELECT id, Val, rn FROM (SELECT id, Val, Col, ROW_NUMBER()OVER (PARTITION BY id ORDER BY Val DESC) AS rn FROM TEMP2 UNPIVOT (Val FOR Col IN ([Week1], [Week2],[Week3],[Week4],[Week5],[Week6],[Week7],[Week8],[Week9], [Week10],[Week11],[Week12],[Week13],[Week14],[Week15],[Week16],[Week17], [Week18],[Week19],[Week20],[Week21],[Week22],[Week23],[Week24],[Week25], [Week26]))AS unpvt) AS t WHERE t.rn <= 6) AS src PIVOT (MAX(Val) FOR rn IN ([1], [2], [3], [4], [5], [6])) AS pvt，就可以先返回最大值。

这对我设置的一些测试数据起了作用：

// myModule.js
import dependency from './dependency';

export default (x) => {
  dependency.doSomething(x * 2);
}

如何查找表

1 个答案: