为我的用户在 SO 上获得徽章的“等级” - 查询速度很慢 - 可以加速吗？

Question

我很好奇有多少人在我之前得到了 - 我能够得到这些信息

python    2019-01-02 09:09:15   Gold    454

有了这个（运行缓慢）query:

（我无法在数据浏览器上与我的主要用户单次/交叉登录，因此匿名登录）

-- insert your user id here:
declare @uid int = 7505395

-- get all badges of all users
select Name, Date, [Gold/Silver/Else], [Row#] from ( 
  SELECT Name, 
         Date, 
         userId,
         case when class = 1 then 'Gold'
              when class = 2 then 'Silver'
              when class = 3 then 'Bronze'
              else convert(varchar(10), class)
              end as 'Gold/Silver/Else',
              ROW_NUMBER() OVER(PARTITION BY name, class ORDER BY date ASC) AS Row# 
  FROM badges
  WHERE 1 = 1
    -- you can restrict this further, f.e. for looking only by gold badges
    -- and Class = 1  -- gold == 1, silver == 2, bronze == 3
    -- -- or for certain named badges
    -- and name like 'python%' 
) as tmp
where userID = @uid 
ORDER by name asc, Date asc

（按原样查询给了我所有的徽章，有多少在我之前得到它，并且必须对所有可能的徽章进行排序）

问题：

我尝试了 CTE（只有错误，没有工作）并且我的 sql 技能生疏了 - 如何加速这个查询？

Answer 1

问题是表格似乎没有对此有用的索引。我们得到如下执行计划：

-- 索引扫描不是最理想的。我们想要索引搜索。

不过，您可以通过以下方式将时间缩短近一半：

1. 预选用户的徽章。
2. 对排名使用相关子查询。
3. 使用Id 作为Date 的代理。 （Ids 是独一无二的，不断增加，而且通常可以更快地进行排序。）

另请注意：

1. the magic ##UserId:INT## parameter的使用。
2. Class 列只有 3 个值。
3. 您可以通过省略ORDER BY 子句将查询时间再缩短几秒钟。

无论如何，this query 表现更好：

WITH zUsersBadges AS (
    SELECT  b.Id
            , b.UserId
            , b.Name
            , b.Date
            , b.Class
            , [Badge Class] = (
                CASE    WHEN b.Class = 1 THEN 'Gold'
                        WHEN b.Class = 2 THEN 'Silver'
                        WHEN b.Class = 3 THEN 'Bronze'
                END
            )
            , [Is tag badge] = IIF (b.TagBased = 1, 'Yes', 'No')
    FROM    Badges b
    WHERE   b.UserId = ##UserId:INT##
)
SELECT      ub.Name                 AS [Badge Name]
            , ub.[Badge Class]
            , ub.[Is tag badge]
            , ub.Date               AS [Date Earned]
            , [In Top N of earners] = (
                SELECT  COUNT (ob.ID)
                FROM    Badges ob
                WHERE   (ob.Name = ub.Name  AND  ob.Class = ub.Class  AND  ob.Id <= Ub.Id)  -- Faster but may give slightly higher rank
                --WHERE   (ob.Name = ub.Name  AND  ob.Class = ub.Class  AND  ob.Date <= Ub.Date)  -- Slower, but gives exact rank.
            )
FROM        zUsersBadges ub
ORDER BY    ub.Name, ub.Date

---

更新：This query 表现更好，因为它聚合了多次获得的徽章：

WITH zUsersBadges AS (
    SELECT      b.UserId
                , b.Name
                , minId = MIN (b.Id)
                , [First Earned] = MIN (b.Date)
                , [Earned N times] = COUNT (b.Date)
                , b.Class
                , [Badge Class] = (
                    CASE    WHEN b.Class = 1 THEN 'Gold'
                            WHEN b.Class = 2 THEN 'Silver'
                            WHEN b.Class = 3 THEN 'Bronze'
                    END
                )
                , [Is tag badge] = IIF (b.TagBased = 1, 'Yes', 'No')
    FROM        Badges b
    WHERE       b.UserId = ##UserId:INT##
    GROUP BY    b.UserId, b.Class, b.Name, b.TagBased
)
SELECT      ub.Name                 AS [Badge Name]
            , ub.[Badge Class]
            , ub.[Is tag badge]
            , ub.[First Earned]
            , ub.[Earned N times]
            , [In Top N of earners] = (
                SELECT  COUNT (ob.ID)
                FROM    Badges ob
                WHERE   (ob.Class = ub.Class  AND  ob.Id <= Ub.minId  AND  ob.Name = ub.Name)  -- Faster but may give slightly higher rank
                --WHERE   (ob.Class = ub.Class  AND  ob.Date <= Ub.[First Earned]  AND  ob.Name = ub.Name)  -- Faster but may give slightly higher rank
            )
FROM        zUsersBadges ub
ORDER BY    ub.Name, ub.[First Earned]

Answer 2

您可以将聚合与过滤一起使用：

select count(*)
from badges b
where b.name = 'python' and b.class = 2 and
      b.date < (select b2.date
                from badges b2
                where b2.name = 'python' and b2.class = 2 and
                      b2.userID = @uid 
               );