如何在SQL中查找所有最常見的值？

Question

我將如何顯示SQL中所有最常見的值？

因此，我在下面顯示了查詢以顯示最常見的值。

SELECT name, COUNT(*) AS popularity
FROM cattwo 
GROUP BY name 
ORDER BY popularity DESC 
LIMIT 1;

    +----------+------------+
    | name     | popularity |
    +----------+------------+
    | cat22610 |          7 |
    +----------+------------+

但是，當我顯示前10個最常見的值時，結果是...

SELECT name, COUNT(*) AS popularity 
FROM cattwo 
GROUP BY name 
ORDER BY popularity DESC 
LIMIT 10;

    +----------+------------+
    | name     | popularity |
    +----------+------------+
    | cat22610 |          7 |
    | cat68704 |          7 |
    | cat14153 |          7 |
    | cat52476 |          7 |
    | cat4556  |          7 |
    | cat64173 |          7 |
    | cat5586  |          7 |
    | cat89302 |          6 |
    | cat97131 |          6 |
    | cat42010 |          6 |
    +----------+------------+

目標是展示所有流行度最高的貓。 這樣的事情。

+----------+------------+
| name     | popularity |
+----------+------------+
| cat22610 |          7 |
| cat68704 |          7 |
| cat14153 |          7 |
| cat52476 |          7 |
| cat4556  |          7 |
| cat64173 |          7 |
| cat5586  |          7 |
+----------+------------+

幫助會很棒。 提前致謝。

Answer 1

SELECT name, COUNT(*) AS popularity 
FROM cattwo 
GROUP BY name 
HAVING COUNT(*) = 
        (
            SELECT COUNT(*) AS popularity 
            FROM cattwo 
            GROUP BY name
            ORDER BY popularity DESC
            LIMIT 1
        )
LIMIT 10;

• SQLFiddle演示

Answer 2

我想這個查詢對您有幫助！

SELECT NAME, COUNT(*) AS POPULARITY 
FROM CATTWO 
GROUP BY NAME 
HAVING COUNT(*) = 
        (
            SELECT COUNT(*) AS MAX_POPULARITY 
            FROM CATTWO 
            GROUP BY NAME
            ORDER BY MAX_POPULARITY DESC
            LIMIT 1
        );

Answer 3

select C1.name,COUNT(*) AS  popularity FROM cattwo C1 GROUP BY C1.name

HAVING 0=(SELECT COUNT(*) popularity  FROM  cattwo 
C2 GROUP BY C2.NAME HAVING C1.popularity <C2.popularity )