如何在SQL中从单行中的多行获取列

我想得到2列col_a和col_b的值，用于其他列的min和max。例如：

身份证件 上次更新 科鲁阿 科鲁布 1. 2021-01-01 abc xyz 1. 2021-01-02 abc_0 xyz_0 1. 2021-01-03 abc_1 xyz_1 1. 2021-01-04 abc_2 xyz_2 2. 2021-01-01 abc xyz 2. 2021-01-01 abc xyz ...

---

对于每个

id

记录组，我们可以使用

ROW\u NUMBER

两次，按

last\u updated

的顺序查找第一行和最后一行。然后，按

id

进行聚合，并为各种

列a

和

列b

值透视出列

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY last_updated) rn_min,
              ROW_NUMBER() OVER (PARTITION BY id ORDER BY last_updated DESC) rn_max
    FROM yourTable
)

SELECT
    id,
    MAX(CASE WHEN rn_min = 1 THEN col_a END) AS col_a_min,
    MAX(CASE WHEN rn_max = 1 THEN col_a END) AS col_a_max,
    MAX(CASE WHEN rn_min = 1 THEN col_b END) AS col_b_min,
    MAX(CASE WHEN rn_max = 1 THEN col_b END) AS col_b_max
FROM cte
GROUP BY id;

---

对于每个

id

记录组，我们可以使用

ROW\u NUMBER

两次，按

last\u updated

的顺序查找第一行和最后一行。然后，按

id

进行聚合，并为各种

列a

和

列b

值透视出列

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY last_updated) rn_min,
              ROW_NUMBER() OVER (PARTITION BY id ORDER BY last_updated DESC) rn_max
    FROM yourTable
)

SELECT
    id,
    MAX(CASE WHEN rn_min = 1 THEN col_a END) AS col_a_min,
    MAX(CASE WHEN rn_max = 1 THEN col_a END) AS col_a_max,
    MAX(CASE WHEN rn_min = 1 THEN col_b END) AS col_b_min,
    MAX(CASE WHEN rn_max = 1 THEN col_b END) AS col_b_max
FROM cte
GROUP BY id;

---

这不是最简单的解决方案，但演示了获取所需数据的另一种方法。我们将表本身连接起来，因为我们通常需要来自两行的数据，然后我们使用

交叉应用

将其限制为第一行和最后一行

select T1.id, T2.col_a, T1.col_a, T2.col_b, T1.col_b
from #my_table T1
inner join #my_table T2 on T1.id = T2.id
cross apply (
  select id, max(last_updated) MaxLastUpdated, min(last_updated) MinLastUpdated
  from #my_table
  group by id
) X
where T1.last_updated = X.MaxLastUpdated and T2.last_updated = X.MinLastUpdated;

---

对于提供的示例数据，这似乎比

行编号（）的解决方案执行得更差。最快的解决方案是分析函数。
不是最简单的解决方案，而是演示了获取所需数据的另一种方法。我们将表本身连接起来，因为我们通常需要来自两行的数据，然后我们使用
交叉应用
将其限制为第一行和最后一行

select T1.id, T2.col_a, T1.col_a, T2.col_b, T1.col_b
from #my_table T1
inner join #my_table T2 on T1.id = T2.id
cross apply (
  select id, max(last_updated) MaxLastUpdated, min(last_updated) MinLastUpdated
  from #my_table
  group by id
) X
where T1.last_updated = X.MaxLastUpdated and T2.last_updated = X.MinLastUpdated;

对于提供的示例数据，这似乎比
行编号（）的解决方案执行得更差。最快的解决方案是分析函数。
您可以使用
MIN
和
MAX
分析函数，如下所示：

select id,
       max(case when mindt = last_updated then col_a end) as min_col_a,
       max(case when maxdt = last_updated then col_a end) as max_col_a,
       max(case when mindt = last_updated then col_b end) as min_col_b,
       max(case when maxdt = last_updated then col_b end) as max_col_b
  from
(select t.*,
       min(last_updated) over (partition by id) as mindt,
       max(last_updated) over (partition by id) as maxdt
  from your_table t) t
group by id

您可以使用
MIN
和
MAX
分析功能，如下所示：

select id,
       max(case when mindt = last_updated then col_a end) as min_col_a,
       max(case when maxdt = last_updated then col_a end) as max_col_a,
       max(case when mindt = last_updated then col_b end) as min_col_b,
       max(case when maxdt = last_updated then col_b end) as max_col_b
  from
(select t.*,
       min(last_updated) over (partition by id) as mindt,
       max(last_updated) over (partition by id) as maxdt
  from your_table t) t
group by id

我如何在单独的列中获取这些内容？我已经更新了问题答案。你最初的问题让人觉得你想要一个CSV字符串。我如何在单独的列中获得这些字符串？我已经更新了问题答案。您最初的问题使您似乎想要一个CSV字符串。