我需要对列执行一个简单的减法运算,并将新值插入到新的table.column中。
表的结构是这样的,逻辑不是一列,我只想在数学上做时间而不是日期,删除它也很好,但日期和时间是在同一个单元格中粘在一起的:

Table A
    A       Time                      (Logic)
1   aaa   YYYY-MM-DD HH:MM:SS         (row2 - row1)
2   aaa   YYYY-MM-DD HH:MM:SS         (row3-row2)
3   aaa   YYYY-MM-DD HH:MM:SS         (row4-row3)


Table B
        A         new_time
    1   aaa       insert logic row 1
    2   aaa       insert logic row 2
    3   aaa       insert logic row 3

问题是如何从下一行减去tableA.time的每一行,并将该值更新为tableab.new_time?
-----编辑----
@克林
这是我模仿你的剧本:
update tableB
set new_time = tableA.time
from (
    select tableA.A- lead(tableA.A) over (order by tableB.B)
    from tableA
    ) tableA.A
where tableB.a = tableA.A

错误消息:
SQL Error [500310] [42P10]: [Amazon](500310) Invalid operation: subquery in FROM may not refer to other relations of same query level;

最佳答案

您可以在update中使用子查询作为数据源。例子:

create table table_a (id int primary key, t timestamp);
insert into table_a values
(1, '2016-09-01'), (2, '2016-09-02'), (3, '2016-09-04');

create table table_b (id int primary key, i interval);
insert into table_b values
(1, null), (2, null), (3, null);

update table_b b
set i = a.i
from (
    select id, t - lead(t) over (order by id) i
    from table_a
    ) a
where b.id = a.id;

select * from table_b;

 id |    i
----+---------
  1 | -1 days
  2 | -2 days
  3 |
(3 rows)

一个字母的别名很方便,但有时不太清楚。具有更多信息别名的同一查询:
update table_b as alias_b
set i = subquery.i
from (
    select id, t - lead(t) over (order by id) i
    from table_a
    ) as subquery
where alias_b.id = subquery.id;

关于sql - 如何执行UPDATE +数学函数为空列?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/39778729/

10-13 04:46