Question

我有这个数据的表格：

 1. John | seafood pizza 
 2. Mike | pepperoni pizza 
 3. Mike | pepperoni pizza
 4. John | original pizza 
 5. Mike | original pizza 
 6. John | seafood pizza
 7. John | pepperoni pizza

...

如何编写一个给我结果的查询：

John | seafood pizza
Mike | pepperoni pizza

如果客户的许多比萨饼的数量相同，结果可能是比萨饼的名称。

Answer 1

让我们一步一步地采取行动

以下查询将为您提供每个客户订购各种披萨的次数

SELECT name, pizza, COUNT(*) AS cnt 
FROM yourTable 
GROUP BY name, pizza

那么我们如何从中获得最常订购的披萨呢？我们必须首先知道，每个客户订购的比萨饼数量是多少

SELECT name, MAX(cnt) AS cnt FROM (
  SELECT name, pizza, COUNT(*) AS cnt 
  FROM yourTable 
  GROUP BY name, pizza
) AS subquery GROUP BY name

然后使用此号码选择实际比萨饼名称

SELECT name, pizza, COUNT(*) AS cnt 
FROM yourTable AS t
CROSS JOIN (
  SELECT name, MAX(cnt) AS cnt FROM (
    SELECT name, pizza, COUNT(*) AS cnt 
    FROM yourTable 
    GROUP BY name, pizza
  ) AS subquery GROUP BY name
) AS sq
USING(name,cnt)
GROUP BY name, pizza

Answer 2

我会按用户和披萨对计数进行分组，然后将其用作查询的数据源，该查询找到groupwise最大值（使用max concat技巧）。这并不难：

SELECT user, SUBSTRING(MAX(CONCAT(LPAD(freq, 6, '0'),pizza)),7)
FROM
(SELECT user, pizza, COUNT(*) AS freq
FROM user_likes
GROUP BY user, pizza) ilv
GROUP BY user

Answer 3

让我们有一个简单的解决方案：首先，我们有两个栏目，名称和食物。您可以轻松列出喜爱的顾客的食物：

CREATE TEMPORARY TABLE tbltemp AS (
SELECT name, food, COUNT( * ) c
FROM food
GROUP BY name, food
);

现在我们有所有的食物。然后你可以通过递减计数来订购它：

SELECT *
FROM tbltemp
ORDER BY c DESC

现在您有一个描述客户最喜欢的列表。

更新：

取代第二个查询，将其替换为：

SELECT tbl1 . *
FROM (

SELECT name, food, COUNT( * ) c
FROM food
GROUP BY name, food
) AS tbl1, (

SELECT name, food, COUNT( * ) c
FROM food
GROUP BY name, food
) AS tbl2
WHERE tbl1.c = tbl2.c
AND tbl2.name = tbl1.name
AND tbl2.food = tbl1.food
AND tbl1.c = (
SELECT c
FROM tbltemp
ORDER BY c DESC
LIMIT 1 )

此查询结果是您的问题目标。

Mysql：哪个披萨更喜欢客户端

3 个答案: