如何用 SQL 将垂直数据转换为水平数据?

How to transform vertical data into horizontal data with SQL?(如何用 SQL 将垂直数据转换为水平数据?)

问题描述

我有一个包含许多相关项目的表格项目"，如下所示:

I have a table "Item" with a number of related items, like so:

ID   Rel_ID  Name  RelRank
---  ------  ----  -------
1    1       foo   1
2    1       bar   2
3    1       zam   3
4    2       foo2  1

我正在尝试获取一个查询，因此具有相同 Rel_ID 的项目将出现在同一行中，如下所示:

I'm trying to get a query so items with the same Rel_ID would appear in the same row, like so:

Rel_ID  Name1  Name2  Name3
------  -----  -----  -----
1       foo    bar    zam
2       foo2

我已尝试多次选择表格:

I've tried selecting the table multiple times:

SELECT k.Rel_ID, k.name 'Name1', k2.name 'Name2'
FROM item k, item k2
WHERE k.Rel_ID = k2.Rel_ID

但这失败了.当然有一个转换或查询可以大大简化这个过程，我只是想念它，因为我以前没有以这种方式使用过 SQL.我错过了什么?

But this fails. Surely there's a transformation or query that could drastically simplify the process, and I'm just missing it because I haven't used SQL in this way before. What am I missing?

推荐答案

不管你使用什么数据库，你想要实现的概念叫做数据透视表".

Regardless of the database you are using, the concept of what you are trying to achieve is called "Pivot Table".

以下是 mysql 的示例:http://en.wikibooks.org/wiki/MySQL/Pivot_table

Here's an example for mysql: http://en.wikibooks.org/wiki/MySQL/Pivot_table

一些数据库具有内置功能，请参阅下面的链接.

Some databases have builtin features for that, see the links below.

SQL服务器:http://msdn.microsoft.com/de-de/library/ms177410.aspx

甲骨文:http://www.dba-oracle.com/t_pivot_examples.htm

您始终可以手动创建枢轴.只需选择结果集中的所有聚合，然后从该结果集中进行选择.请注意，在您的情况下，您可以使用 concat 将所有名称放入一列(我认为这是 mysql 中的 group_concat)，因为您不知道有多少名称与 rel_id 相关.

You can always create a pivot by hand. Just select all the aggregations in a result set and then select from that result set. Note, in your case, you can put all the names into one column using concat (i think that's group_concat in mysql), since you cannot know how many names are related to a a rel_id.

为您的案例伪选择(我不知道 mysql):

pseudo-select for your case (i don't know mysql):

select rel_id, group_concat(name) from item group by rel_id

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