SQL Server LEAD 函数答案

【问题标题】：SQL Server LEAD functionSQL Server LEAD 函数
【发布时间】：2021-10-28 15:19:08
【问题描述】：

-- FIRST LOGIN DATE
WITH CTE_FIRST_LOGIN AS 
(
    SELECT 
        PLAYER_ID, EVENT_DATE, 
        ROW_NUMBER() OVER (PARTITION BY PLAYER_ID ORDER BY EVENT_DATE ASC) AS RN
    FROM 
        ACTIVITY
),
-- CONSECUTIVE LOGINS
CTE_CONSEC_PLAYERS AS 
(
    SELECT 
        PLAYER_ID, 
        LEAD(EVENT_DATE,1) OVER (PARTITION BY EVENT_DATE ORDER BY EVENT_DATE) NEXT_DATE 
    FROM 
        ACTIVITY A
    JOIN 
        CTE_FIRST_LOGIN C ON A.PLAYER_ID = C.PLAYER_ID
    WHERE  
        NEXT_DATE = DATEADD(DAY, 1, A.EVENT_DATE) AND C.RN = 1
    GROUP BY 
        A.PLAYER_ID
)
-- FRACTION
SELECT 
    NULLIF(ROUND(1.00 * COUNT(CTE_CONSEC.PLAYER_ID) / COUNT(DISTINCT PLAYER_ID), 2), 0) AS FRACTION 
FROM 
    ACTIVITY 
JOIN 
    CTE_CONSEC_PLAYERS CTE_CONSEC ON CTE_CONSEC.PLAYER_ID = ACTIVITY.PLAYER_ID

我在运行此查询时收到以下错误。

[42S22] [Microsoft][ODBC Driver 17 for SQL Server][SQL Server]列名“NEXT_DATE”无效。 (207) (SQLExecDirectW)

这是一个 leetcode 中等问题 550。游戏玩法分析 IV。我想知道为什么它不能在这里识别列 NEXT_DATE 以及我错过了什么？谢谢！

【问题讨论】：

标签： sql sql-server-2012 lead

【解决方案1】：

问题出在这个 CTE 上：

-- CONSECUTIVE LOGINS prep
CTE_CONSEC_PLAYERS AS (
  SELECT 
    PLAYER_ID, 
    LEAD(EVENT_DATE,1) OVER (PARTITION BY EVENT_DATE ORDER BY EVENT_DATE) NEXT_DATE 
  FROM ACTIVITY A
  JOIN CTE_FIRST_LOGIN C  ON A.PLAYER_ID = C.PLAYER_ID
  WHERE  NEXT_DATE = DATEADD(DAY, 1, A.EVENT_DATE) AND C.RN = 1
  GROUP BY A.PLAYER_ID
)

请注意，您正在创建 NEXT_DATE 作为此 CTE 中的列别名，但也在 WHERE 子句中引用它。这是无效的，因为根据 SQL 子句排序规则，NEXT_DATE 列别名不存在，直到您到达 ORDER BY 子句，这是 SQL 查询或子查询中的最后一个评估子句。您在此子查询中没有 ORDER BY 子句，因此从技术上讲，NEXT_DATE 列别名仅存在于之后并引用您的 CTE_CONSEC_PLAYERS CTE 的 [子] 查询。

要解决此问题，您可能需要两个这样的 CTE（未经测试）：

-- CONSECUTIVE LOGINS
CTE_CONSEC_PLAYERS_pre AS (
  SELECT 
    PLAYER_ID, 
    RN,
    EVENT_DATE,
    LEAD(EVENT_DATE,1) OVER (PARTITION BY EVENT_DATE ORDER BY EVENT_DATE) NEXT_DATE 
  FROM ACTIVITY A
  JOIN CTE_FIRST_LOGIN C  ON A.PLAYER_ID = C.PLAYER_ID
)
-- CONSECUTIVE LOGINS
CTE_CONSEC_PLAYERS AS (
  SELECT
    PLAYER_ID, 
    MAX(NEXT_DATE) AS NEXT_DATE,
  FROM CTE_CONSEC_PLAYERS_pre
  WHERE  NEXT_DATE = DATEADD(DAY, 1, EVENT_DATE) AND RN = 1
  GROUP BY PLAYER_ID
)

【讨论】：

【解决方案2】：

您给每个表一个别名（例如JOIN CTE_FIRST_LOGIN C 具有别名C），并且每个列访问都是通过别名。您需要将正确表中的正确别名添加到NEXT_DATE。

【讨论】：

【解决方案3】：

您的主要问题是 NEXT_DATE 是一个窗口函数，因此由于 SQL 的操作顺序，不能在 WHERE 中引用。

但这个查询似乎过于复杂了。

要解决的问题似乎是：有多少玩家在首次登录后的第二天登录，占所有玩家的百分比。

这可以通过同时使用多个窗口函数在单次传递（无连接）中完成：

WITH CTE_FIRST_LOGIN AS (
    SELECT
      PLAYER_ID,
      EVENT_DATE,
      ROW_NUMBER() OVER (PARTITION BY PLAYER_ID ORDER BY EVENT_DATE) AS RN,
-- if EVENT_DATE is a datetime and can have multiple per day then group by CAST(EVENT_DATE AS date) first
      LEAD(EVENT_DATE, 1) OVER (PARTITION BY EVENT_DATE ORDER BY EVENT_DATE) AS NextDate
  FROM ACTIVITY
),
BY_PLAYERS AS (
    SELECT
      c.PLAYER_ID,
      SUM(CASE WHEN c.RN = 1 AND c.NextDate = DATEADD(DAY, 1, c.EVENT_DATE)
        THEN 1 END) AS IsConsecutive
    FROM CTE_FIRST_LOGIN AS c
    GROUP BY c.PLAYER_ID
)
SELECT ROUND(
    1.00 *
    COUNT(c.IsConsecutive) /
    NULLIF(COUNT(*), 0)
  ,2) AS FRACTION
FROM BY_PLAYERS AS c;

理论上，您可以将BY_PLAYERS 合并到外部查询中并使用COUNT(DISTINCT，但拆分它们感觉更干净

【讨论】：