MySQL为每个类别选择前10个项目[重复]

Question

您好，我有一个 mysql 数据库，其中有多个类别。例如，我希望能够仅提取每个类别的前 10 个项目：

我有下表，我想为 name = a 提取前 2 行，name = b 和 name = c 相同

-----------------------------------
name | value | description | logo
-----------------------------------
a    | 2.00  | its a letter| image
-----------------------------------
a    | 5.00  | its a letter| image
-----------------------------------
b    | 6.00  | its a letter| image
-----------------------------------
c    | 3.00  | its a letter| image
-----------------------------------
c    | 1.00  | its a letter| image
------------------------------------

这就是我所拥有的，所以农场后过滤器是调用时传入的一串对象。不幸的是，它只给了我所有东西的前 10 个，如果你能指出我正确的方向，那就太好了，谢谢！

代码：

SELECT * 
FROM object_list 
Where object IN (".$_POST['filter'].") 
ORDER BY date DESC, object ASC,id DESC

Answer 1

您可以通过以下查询获取组以及元素count：

SELECT name, value, 
@count := IF(@value = name, @count + 1, 1) AS count,
@value := name AS some_value
FROM test, (SELECT @count := 1, @value := NULL) a
WHERE test.name in ('a', 'b')

现在，如果您需要将每组的行数限制为 2，那么您只需将此查询包装到另一个选择中并添加一个条件，例如：

SELECT * 
FROM (
  SELECT name, value, 
  @count := IF(@value = name, @count + 1, 1) AS count,
  @value := name AS some_value
  FROM test, (SELECT @count := 1, @value := NULL) a
  WHERE test.name in ('a', 'b')) a
WHERE a.count <= 2;

这里是 SQL Fiddle。

Answer 2

这行得通吗？

SELECT 
    yourtable.* 
FROM 
    yourtable 
    JOIN (
        SELECT 
            t1.name, 
            t1.value, 
            COUNT(t2.name) AS theCount 
        FROM yourtable t1 
        LEFT JOIN yourtable t2 ON t1.name = t2.name AND t1.value > t2.value 
        WHERE t1.name in ('a', 'b')
        GROUP BY t1.name, t1.value 
        HAVING theCount < 2
    ) AS dt USING (name, value);

来源：http://thenoyes.com/littlenoise/?p=36