Question

例如，我有2个时间表： T1

id time
1 18:12:02
2 18:46:57
3 17:49:44
4 12:19:24
5 11:00:01
6 17:12:45

和T2

id time
1 18:13:02
2 17:46:57

我需要从T1获得与T2时间最接近的时间。这些表之间没有关系。它应该是这样的：

select T1.calldatetime
from T1, T2 
where T1.calldatetime between 
T2.calldatetime-(
    select MIN(ABS(T2.calldatetime-T1.calldatetime))
    from T2, T1)
and
T2.calldatetime+(
    select MIN(ABS(T2.calldatetime-T1.calldatetime))
    from T2, T1)

但我无法得到它。有什么建议吗？

Answer 1

与使用多个解决方案的其他解决方案不同，您只需使用单个笛卡尔联接来解决问题。我假设时间存储为VARCHAR2。如果它存储为日期，则可以删除TO_DATE函数。如果它存储为日期（我强烈推荐这个），你将不得不删除日期部分

我已经略显冗长，所以很明显发生了什么。

select *
  from ( select id, tm
              , rank() over ( partition by t2id order by difference asc ) as rnk
           from ( select t1.*, t2.id as t2id
                       , abs( to_date(t1.tm, 'hh24:mi:ss') 
                              - to_date(t2.tm, 'hh24:mi:ss')) as difference
                    from t1
                   cross join t2
                         ) a
                 )
 where rnk = 1

基本上，这可以计算出T1和T2中每次的绝对差值，然后选择T2 ID的最小差值;从T1返回数据。

此处位于SQL Fiddle format。

不太漂亮（但更短）的格式是：

select *
  from ( select t1.*
              , rank() over ( partition by t2.id 
                                  order by abs(to_date(t1.tm, 'hh24:mi:ss') 
                                            - to_date(t2.tm, 'hh24:mi:ss'))
                                  ) as rnk
           from t1
          cross join t2
                ) a
 where rnk = 1

Answer 2

我相信这是您正在寻找的查询：

CREATE TABLE t1(id INTEGER, time DATE);
CREATE TABLE t2(id INTEGER, time DATE);

INSERT INTO t1 VALUES (1, TO_DATE ('18:12:02', 'HH24:MI:SS'));
INSERT INTO t1 VALUES (2, TO_DATE ('18:46:57', 'HH24:MI:SS'));
INSERT INTO t1 VALUES (3, TO_DATE ('17:49:44', 'HH24:MI:SS'));
INSERT INTO t1 VALUES (4, TO_DATE ('12:19:24', 'HH24:MI:SS'));
INSERT INTO t1 VALUES (5, TO_DATE ('11:00:01', 'HH24:MI:SS'));
INSERT INTO t1 VALUES (6, TO_DATE ('17:12:45', 'HH24:MI:SS'));

INSERT INTO t2 VALUES (1, TO_DATE ('18:13:02', 'HH24:MI:SS'));
INSERT INTO t2 VALUES (2, TO_DATE ('17:46:57', 'HH24:MI:SS'));

SELECT t1.*, t2.*
  FROM t1, t2,
       (  SELECT t2.id, MIN (ABS (t2.time - t1.time)) diff
            FROM t1, t2
        GROUP BY t2.id) b
 WHERE ABS (t2.time - t1.time) = b.diff;

确保时间列具有相同的日期部分，因为t2.time - t1.time部分不起作用。

编辑：感谢您的接受，但Ben的回答更好。它使用Oracle分析功能，并且性能会更好。

Answer 3

这里的这一行从T1中选择行，其与T2中的任何行距离最小：

select T1.id, T1.calldatetime from T1, T2 
 where ABS(T2.calldatetime-T1.calldatetime)
         =( select MIN(ABS(T2.calldatetime-T1.calldatetime))from T1, T2);

（用mysql测试它，希望你不要从中获得ORA）

编辑：根据上一条评论，它应该是这样的：

drop table t1;
drop table t2;
create table t1(id int, t time);
create table t2(id int, t time);

insert into t1 values (1, '18:12:02');
insert into t1 values (2, '18:46:57');
insert into t1 values (3, '17:49:44');
insert into t1 values (4, '12:19:24');
insert into t1 values (5, '11:00:01');
insert into t1 values (6, '17:12:45');

insert into t2 values (1, '18:13:02');
insert into t2 values (2, '17:46:57');

select ot2.id, ot2.t, ot1.id, ot1.t from t2 ot2, t1 ot1  
 where ABS(ot2.t-ot1.t)=
       (select min(abs(t2.t-t1.t)) from t1, t2 where t2.id=ot2.id)

产地：

id      t       id      t
1       18:13:02        1       18:12:02
2       17:46:57        3       17:49:44

Answer 4

使用分析函数的另一种方法。可能很奇怪:)）

select id, time,
case 
  when to_date(time, 'hh24:mi:ss') - to_date(lag_time, 'hh24:mi:ss') < to_date(lead_time, 'hh24:mi:ss') - to_date(time, 'hh24:mi:ss')
    then lag_time 
    else lead_time
  end closest_time
from (
select id, tbl,
  LAG(time, 1, null) OVER (ORDER BY time) lag_time, 
  time,
  LEAD(time, 1, null) OVER (ORDER BY time) lead_time 
from
  (
  select id, time, 1 tbl from t1
  union all
  select id, time, 2 tbl from t2
  )
)
where tbl = 2

至SQLFiddle ......及以后！

Answer 5

尝试这个查询有点冗长，我会尝试优化它

select * from t1
where id in (
select id1 from 
(select id1,id2,
rank()  over (partition by id2 order by diff) rnk
from
(select distinct t1.id id1,t2.id id2,
round(min(abs(to_date(t1.time,'HH24:MI:SS') - to_date(t2.time,'HH24:MI:SS'))),2) diff 
from 
t1,t2
group by t1.id,t2.id) )
where rnk = 1);

如何在Oracle sql中获取最接近的日期

5 个答案: