Oracle SQL - 使用 LAG 或 LEAD 将多行结果返回到一行中

Question

我想弄清楚如何 return 1 行获得多行结果。

目前我的代码如下所示：

select x.Reference,
x.date "Date1",
x.char "Char1",
lead(x.date, 1) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Date2",
lead(x.char, 1) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Char2",
lead(x.date, 2) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Date3",
lead(x.char, 2) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Char3"

from tbl x

table 每个 x.Reference 都有多个条目。对于每个 x.Reference 的第一行，行 return 是所需的结果。但是，正如预期的那样，它会继续为找到的每个 x.reference 输出一行。我试图找到一种方法来限制这些额外行的输出，因为第一行已经有我需要的数据。

提前致谢。

Answer 1

嗯，你可以使用 row_number():

with x as (
      select x.Reference, x.date as "Date1", x.char as "Char1",
             lead(x.date, 1) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Date2",
             lead(x.char, 1) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Char2",
             lead(x.date, 2) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Date3",
             lead(x.char, 2) OVER (PARTITION BY x.Reference ORDER BY x.date) as "Char3",
             row_number() over (partition by x.Reference order by x.date) as seqnum
      from tbl x
     )
select x.*
from x
where seqnum = 1;

Oracle SQL - 使用 LAG 或 LEAD 将多行结果返回到一行中

Oracle SQL - Returning multiple row results into one single row using LAG or LEAD

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lag

lead