Php 我如何进行此查询？

这是测试表

----------------------------------------------
id | username | point | level | created_date
----------------------------------------------
 1 | name_a   |  1    |   1   | 2011-08-01
 2 | name_a   |  2    |   2   | 2011-08-02
 3 | name_b   |  5    |   1   | 2011-08-02
 3 | name_c   |  6    |   1   | 2011-08-02
 4 | name_d   |  1    |   1   | 2011-08-01
 5 | name_d   |  3    |   1   | 2011-08-02
 5 | name_d   |  5    |   2   | 2011-08-03
 4 | name_e   |  5    |   1   | 2011-08-01
 5 | name_e   |  5    |   2   | 2011-08-02
 5 | name_e   |  5    |   3   | 2011-08-03
----------------------------------------------

查询的要求是在一次查询中尽可能多地查询表的用户名和点

按每个级别的用户分数总和排序。 如果用户在同一级别上有两个分数，它将只获得最新的分数。 按用户名分组 总分必须小于10分 每个级别的最高分数为5分 输出样本：

--------------------
 username  | tpoint|
--------------------
  name_d   |  8    |
  name_b   |  5    |
  name_a   |  3    |
--------------------

---

name_e和name_c被忽略。

听起来像是一个有趣的查询

SELECT username, SUM(point) AS points
FROM (SELECT username, level, point
      FROM (SELECT username, level, LEAST(point, 5) AS point
            FROM table
            WHERE points <= 5
            ORDER BY created_date DESC) AS h
      GROUP BY username, level) AS h2
GROUP BY username
HAVING points < 10
ORDER BY points DESC

这应该够了！换张桌子就行了

编辑：

是否要排除得分超过5的行，或将值设置为5？只需删除其中的点 不知道我是否正确使用了别名，希望这能奏效

---

连接的内部查询是获取最新的用户名、级别组合，其中单个级别点计数大于5。然后，这将用于获得每个用户名的总积分，并丢弃分数超过10分的用户名。

好的，先参加第2部分

SELECT *
FROM table a
WHERE NOT EXISTS (
   SELECT 1
   FROM table b
   WHERE b.username=a.username
   AND a.created_date>b.created_date
)

但是mysql不能很好地处理推送谓词，因此使用了max-concat技巧，但是这将使查询变得非常复杂——如果您遇到性能问题，那么值得重新访问。现在加入其他内容…第1、3和5部分

SELECT username, level, SUM(point)
FROM 
(SELECT *
   FROM table a
   WHERE NOT EXISTS (
      SELECT 1
      FROM table b
      WHERE b.username=a.username
      AND a.created_date>b.created_date
   )
) ilv
GROUP BY username, level
HAVING SUM(point) <= 5;

如何实现4取决于此约束相对于其他约束（特别是2和5）应用的确切顺序。以下应给出所述输入的disred输出

SELECT username, level, SUM(point)
FROM 
(SELECT *
   FROM table a
   WHERE NOT EXISTS (
      SELECT 1
      FROM table b
      WHERE b.username=a.username
      AND a.created_date>b.created_date
   )
) ilv,
(SELECT username, SUM(point) as totpoint
  FROM table c
  GROUP BY username
  HAVING SUM(point)<=10) ilv2
WHERE ilv.username=ilv2.username
GROUP BY username, level
HAVING SUM(point) <= 5;

---

哎哟-再看一遍，你不想看到输出集中的级别细分-在这种情况下，Robin的答案更好。

我认为这一点很接近，但你需要添加一个case语句来检查分数。它首先获取一个表，并对创建的日期进行排序。然后它根据用户名和级别进行分组，这将给point提供最新的值，按created_date DESC排序。您需要第二次选择吗？此外，仅在需要时使用select*。非常好！如果您在查询中发现任何其他值得更改的内容，请告诉我！：

SELECT 
    Query2.username 
  , Sum(Query2.SomVanpoint) AS point 
FROM 
    (SELECT 
            test.username 
          , test.level 
          , Sum(test.point) AS SomVanpoint 
        FROM 
            test 
        INNER 
        JOIN 
            (SELECT 
                    test.username 
                  , test.level 
                  , Max(test.created_date) AS MaxVancreated_date 
                FROM 
                    test 
                GROUP 
                    BY test.username 
                  , test.level
            ) AS Query1 
            ON 
            (test.username         = Query1.username) 
            AND (test.level        = Query1.level) 
            AND (test.created_date = Query1.MaxVancreated_date) 
        GROUP 
            BY test.username 
          , test.level 
        HAVING 
            (((Sum(test.point))<= 5))
    ) AS Query2 
GROUP 
    BY Query2.username 
HAVING 
    (((Sum(Query2.SomVanpoint))< 10)) 
ORDER 
    BY Sum(Query2.SomVanpoint) DESC; 

SELECT 
    Query2.username 
  , Sum(Query2.SomVanpoint) AS point 
FROM 
    (SELECT 
            test.username 
          , test.level 
          , Sum(test.point) AS SomVanpoint 
        FROM 
            test 
        INNER 
        JOIN 
            (SELECT 
                    test.username 
                  , test.level 
                  , Max(test.created_date) AS MaxVancreated_date 
                FROM 
                    test 
                GROUP 
                    BY test.username 
                  , test.level
            ) AS Query1 
            ON 
            (test.username         = Query1.username) 
            AND (test.level        = Query1.level) 
            AND (test.created_date = Query1.MaxVancreated_date) 
        GROUP 
            BY test.username 
          , test.level 
        HAVING 
            (((Sum(test.point))<= 5))
    ) AS Query2 
GROUP 
    BY Query2.username 
HAVING 
    (((Sum(Query2.SomVanpoint))< 10)) 
ORDER 
    BY Sum(Query2.SomVanpoint) DESC; 

username  | point
----------+------
name_d    |  8
name_b    |  5
name_a    |  3