如何为每一行将 DateTime 列增加一秒? [英] How to increment DateTime column by a second for every row?

问题描述

假设我有这个时间值:09:00:00

我有一个包含时间列的表，其中有 3 条记录.

And I have a table with a column Time, and I have three records in it.

我想用那个时间更新这 3 条记录，但 Time 值每次增加一秒(对于每条记录).

I want to update those 3 records with that time but with the Time value incremented in one second each time (for every record).

像这样:

ColumnA   ColumnB
1         09:00:00
2         09:00:01
3         09:00:02

我该怎么做?

我的解决方案:

在我自己的解决方案上工作了一段时间后，这就是我想出的

After some time working on my own solution, this is what I came up with

update tor1
set ColumnB = dateadd(s,tor2.inc, ColumnB)
from table1 tor1
inner join (select ColumnA, ROW_NUMBER() OVER (Order by ColumnA) as inc from table1) tor2 on tor1.ColumnA=tor2.ColumnA

推荐答案

您没有指定任何特定的顺序.

You don't specify any particular ordering.

对于不确定/未记录的结果，您可以尝试古怪的更新方法.

For undeterministic/undocumented results you could try the quirky update approach.

CREATE TABLE table1
(
    ColumnB datetime NULL
);

INSERT INTO table1 DEFAULT VALUES;
INSERT INTO table1 DEFAULT VALUES;
INSERT INTO table1 DEFAULT VALUES;

DECLARE @ColumnB datetime;

SET @ColumnB = '19000101 09:00:00';

UPDATE table1
SET @ColumnB = ColumnB = DATEADD(s, 1, @ColumnB);

SELECT *
FROM table1;

DROP TABLE table1;

否则您将需要使用游标或找到某种方法来模拟 2000 年的 ROW_NUMBER.

Otherwise you will need to use a cursor or find some way of simulating ROW_NUMBER in 2000.

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