Question

我确信必须有一种相对简单的方法来做到这一点，但它现在正在逃避我。假设我有一个这样的SQL表：

+-----+-----+-----+-----+-----+
|  A  |  B  |  C  |  D  |  E  |
+=====+=====+=====+=====+=====+
|  1  |  2  |  3  | foo | bar | << 1,2
+-----+-----+-----+-----+-----+
|  1  |  3  |  3  | biz | bar | << 1,3
+-----+-----+-----+-----+-----+
|  1  |  2  |  4  |  x  |  y  | << 1,2
+-----+-----+-----+-----+-----+
|  1  |  2  |  5  | foo | bar | << 1,2
+-----+-----+-----+-----+-----+
|  4  |  2  |  3  | foo | bar | << 4,2
+-----+-----+-----+-----+-----+
|  1  |  3  |  3  | foo | bar | << 1,3
+-----+-----+-----+-----+-----+

现在，我想知道列A和B的每个值组合出现多少次，而不管其他列。所以，在这个例子中，我想要一个像这样的输出：

+-----+-----+-----+
|  A  |  B  |count|
+=====+=====+=====+
|  1  |  2  |  3  |
+-----+-----+-----+
|  1  |  3  |  2  |
+-----+-----+-----+
|  4  |  2  |  1  |
+-----+-----+-----+

确定这个的SQL是什么？我觉得这绝不是一件非常不寻常的事情。

谢谢！

Answer 1

SELECT A,B,COUNT(*)
FROM the-table
GROUP BY A,B

Answer 2

TRY：

SELECT
    A, B , COUNT(*)
    FROM YourTable
    GROUP BY A, B

Answer 3

这应该这样做：

SELECT A, B, COUNT(*) 
FROM TableName
GROUP BY A, B;

Answer 4

SELECT A,B,COUNT(1) As COUNT_OF
FROM YourTable
GROUP BY A,B

Answer 5

SELECT A,B,COUNT(*)
FROM table
GROUP BY A,B

Answer 6

SELECT A，B，COUNT（*）来自MyTable GROUP BY A，B

Answer 7

这可能就是答案：

SELECT a, b, COUNT(*) 
FROM <your table name here> 
GROUP BY a,b 
ORDER BY 3 DESC;

如何计算两列中具有相同值的行（SQL）？

7 个答案: