Question

我的桌子看起来像这样：

+------+------+-------+------+
| ColA | ColB | ColC  | ColD |
+------+------+-------+------+
| 1    | 1    | 10217 | 5    |
+------+------+-------+------+
| 1    | 2    | 10217 | 6    |
+------+------+-------+------+
| 1    | 1    | 10417 | 7    |
+------+------+-------+------+
| 1    | 2    | 10417 | 8    |
+------+------+-------+------+
| 2    | 1    | 10417 | 9    |
+------+------+-------+------+
| 2    | 2    | 10417 | 10   |
+------+------+-------+------+

如何编写一个查询，该查询将返回一对具有以下要求的行：

输入ColA值和ColC范围作为参数后

例如ColA: 1 ColC: 10217 to 10417

该表应如下所示：

+------+------+-------+------+
| ColA | ColB | ColC  | ColD |
+------+------+-------+------+
| 1    | 1    | 10217 | 5    |
+------+------+-------+------+
| 1    | 2    | 10217 | 6    |
+------+------+-------+------+
| 1    | 1    | 10417 | 7    |
+------+------+-------+------+
| 1    | 2    | 10417 | 8    |
+------+------+-------+------+

现在为特定的ColC值添加每个ColB配对的ColD值。在我的示例中，对于ColC值10217，它将是5 + 6，而ColC值10417将是7 + 8。
然后获得两个和之间的最大值。所以应该是15> 11。
最后，返回由于获得上述最大值（即15）而导致的行配对。

最终结果应如下所示：

+------+------+-------+------+
| ColA | ColB | ColC  | ColD |
+------+------+-------+------+
| 1    | 1    | 10417 | 7    |
+------+------+-------+------+
| 1    | 2    | 10417 | 8    |
+------+------+-------+------+

我只得到最大的部分。我在尝试获取行配对时遇到了麻烦。

SELECT Max(sumD) AS maxSumD
FROM (SELECT t1.ColA, t1.ColB, sum(ColD) AS sumD 
        FROM (SELECT * 
                FROM Table1 
                WHERE (ColA = paramA) and (ColC between paramFromC and paramToC))  AS t1 
        GROUP BY t1.ColA, t1.ColB)

解决方案：使用Ctznkane的答案，我找到了这个（棘手的）解决方案：

SELECT TOP 2 SUMMEDVALUE, ColA, ColB, ColC, ColD
FROM (SELECT A.ColA, A.ColB, A.ColC, A.ColD, (A.ColD+B.ColD) as SUMMEDVALUE
      FROM Table1 A
      INNER JOIN Table1 B 
      ON A.ColA=B.ColA AND A.ColC=B.ColC AND t1.ColB<>t2.ColB) 
ORDER BY SUMMEDVALUE DESC

Answer 1

如果要查找一对行，则应该将表与其自身连接起来。

SELECT (A.COLD+B.COLD) SUMMEDVALUE, A.COLC, A.COLB, B.COLB, A.COLD, B.COLD
FROM TABLE1 A
    INNER JOIN TABLE1 B ON A.COLA=B.COLA AND A.COLC=B.COLC AND A.COLB<>B.COLB

SQL-根据一列中的值的总和获得一对行

1 个答案: