字段少于 select 的分组依据

Question

我有两个 Oracle 表：

USER(ID*,NAME,SURNAME)
MATCH(ID*,START_DATE,END_DATE,MATCH_CODE,ID_USER**)

我需要一个查询来为每个 USER 获取 END_DATE 和 START_DATE 以及 NAME 和 MATCH_CODE 字段之间以秒为单位的最大差异的匹配项。

我的查询：

SELECT A.ID,A.NAME,MAX(extract(second from (END_DATE-START_DATE))
                     + extract(minute from (END_DATE-START_DATE)*60
                     + extract(hour from (END_DATE-START_DATE)*60*60
                     + extract(day from (END_DATE-START_DATE)*60*60*24) max_differance
FROM USER A JOIN MATCH B
ON A.ID = B.ID_USER
GROUP BY A.ID;

我正在考虑这个查询，但显然它给出了一个错误，因为在 GROUP BY 中 select 的所有字段都去。另外我需要 MATCH_CODE 字段，我该怎么办？

Answer 1

聚合 name 列并使用 MAX ... KEEP 得到 match_code:

SELECT u.id,
       MAX(u.name) AS name,
       MAX(end_date - start_date)*24*60*60 AS max_difference,
       MAX(match_code) KEEP (
         DENSE_RANK LAST
         ORDER BY end_date - start_date NULLS FIRST
       ) As match_code
FROM   "USER" u
       INNER JOIN match m
       ON (u.id = m.id_user)
GROUP BY u.id

或者，使用解析函数：

SELECT id,
       name,
       max_difference,
       match_code
FROM   (
  SELECT u.id,
         u.name,
         (end_date - start_date)*24*60*60 AS max_difference,
         match_code,
         ROW_NUMBER() OVER (PARTITION BY u.id ORDER BY end_date - start_date DESC)
           AS rn
  FROM   "USER" u
         INNER JOIN match m
         ON (u.id = m.id_user)
)
WHERE  rn = 1;

字段少于 select 的分组依据

Group By with fewer fields than the select

sql

oracle

greatest-n-per-group