如何計算兩列中具有相同值的行（SQL）？

Question

我相信肯定有一個相對直接的方法來做到這一點，但目前它正在逃避我。假設我有一個SQL表是這樣的：

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

現在，我想知道的列值A和B的每個組合有多少次出現，而不管其他列。所以，在這個例子中，我想要一個這樣的輸出：

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

什麼是SQL來確定？我覺得這不是一件非常不尋常的事情要做。

謝謝！

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

這應做到：

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

Answer 4

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

Answer 5

選擇A，B，COUNT（*） FROM MyTable的 GROUP BY A，B

Answer 6

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

Answer 7

這可能是答案：

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