快速将列的副本添加到MySQL表

I need a fast way duplicate a DATETIME column in a table and give it a new name.

我需要快速复制表中的DATETIME列并为其指定新名称。

I have a column named myDate in my table called myResults, I need a query to make a new column in the table called newDate which has the exact same data as the myDate column.

我的表中有一个名为myRate的名为myDate的列，我需要一个查询来在表中创建一个名为newDate的新列，该列与myDate列具有完全相同的数据。

Is there a faster way to do this than by doing the obvious 2 step approach of make a new column, and then copying all the data (it's a large table and I'm looking for the fastest approach)?

有没有比通过制作新列的明显的两步方法更快的方法，然后复制所有数据（这是一个大表，我正在寻找最快的方法）？

Obvious solution:

明显的解决方案：

ALTER TABLE `myResults` ADD `newDate` DATETIME;  
UPDATE `myResults` SET `newDate` = `myDate`;

4 个解决方案

#1

The obvious solution is the only solution, unfortunately.

不幸的是，明显的解决方案是唯一的解决方案。

However note that in general you shouldn't be copying a column in relational databases.

但请注意，通常不应复制关系数据库中的列。

#2

UPDATE `table_name` SET `new_column` = `existing_column` WHERE `id`=`id`

#3

Why would your workload ever demand a new datetime column, that duplicates another columns data? This sounds like horrable practice? How about telling us what you're trying to achieve? You can pull a second column with the same data in a few different ways, without actually duplicating the data:

为什么您的工作负载需要一个新的日期时间列，它会复制另一个列数据？这听起来像是可怕的练习？告诉我们你想要达到的目标怎么样？您可以通过几种不同的方式使用相同的数据提取第二列，而无需实际复制数据：

SELECT date1 AS date_old, date1 AS date_new FROM table;

-or-, you can create a view

- 或 - ，您可以创建一个视图

CREATE VIEW virtual_table AS
    SELECT date1 AS date_old, date1 AS date_new FROM table
;
SELECT * FROM virtual_table;

#4

If you just need a default in there, you can either choose what the default is statically or use a function call.

如果您只需要默认值，则可以选择静态默认值或使用函数调用。

ALTER TABLE `myResults` ADD `newDate` DATETIME DEFAULT '2010-01-01';

要么

ALTER TABLE `myResults` ADD `newDate` DATETIME DEFAULT current_timestamp;

#1