SQL：获取日期最少的行 + 在 select 语句中使用 CASE WHEN

Question

我有一个包含 3 列的 table (id, type, date) 具有以下值：

id    type    date
---   ---      ---
191     0    2016-09-15 11:26:51.000
191     1    2016-09-15 11:30:31.000
200     0    2016-09-15 17:36:19.000
200     1    2016-09-15 18:26:51.000
331     0    2016-09-16 07:26:22.000

这就是我想要做的：

对于上面的每个 ticketid，select 最小日期最小的行。完成后，我希望我的输出中有另一列名为 "isEngaged"，如果 ticketid 的任何记录都存在 type = 1，则该列将设置为 1。

我想要的输出：

id    type    date                        isEngaged
---   ---      ---                         ---
191     0    2016-09-15 11:26:51.000        1
200     0    2016-09-15 17:36:19.000        1
331     0    2016-09-16 07:26:22.000        0

这是我目前所拥有的：（这只负责返回日期最少的行）

SELECT id, type, date FROM (SELECT ROW_NUMBER() OVER (PARTITION BY id ORDER BY date ASC) AS [Row], id, type, date)
      FROM mytable WHERE type IN (0)) WHERE [Row] = 1

真的不确定如何合并 isEngaged 部分。我尝试使用 CASE WHEN 但不确定如何继续：

SELECT id, type, date,
"isEngaged" = CASE WHEN (SELECT * FROM mytable where type in (1) and id = <not sure what to put here>) THEN 1 ELSE 0 END
FROM mytable

如果您知道更好的方法，请告诉我。

Answer 1

select
id,
case when sum_engaged > 0 then 1 else 0 end as isEngaged,
min_date
from
(SELECT id, 
sum(case when type IN( 1,3) then 1 else 0 end) as sum_engaged, min(date) as min_date 
FROM mytable 
group by id) g

Answer 2

如果您只接受 type 列中的 0/1 值，那么您可以将 MAX() 用作 window 函数。

SELECT  
  id, type, date, isEngaged
FROM (
  SELECT
    id, type, date,
    MAX(type) OVER (PARTITION BY id) AS isEngaged
    ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
  FROM mytable
  ) t
WHERE rn = 1

如果您接受大于 1 的值，那么 MAX() 中适当的 CASE 语句就足够了：

SELECT  
  id, type, date, isEngaged
FROM (
  SELECT
    id, type, date,
    MAX(CASE WHEN type = 1 THEN 1 ELSE 0 END) OVER (PARTITION BY id) AS isEngaged
    ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
  FROM mytable
  ) t
WHERE rn = 1

SQL：获取日期最少的行 + 在 select 语句中使用 CASE WHEN

SQL: Fetch rows with least date + Use CASE WHEN with select statement

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