SQL Server 将多行合并为一行多列

问题描述

我有一个数据库，其中包含以下行:

I have a database in which I have the following rows:

  ID  |  Date start  |  Date end
----------------------------------
  a   |  01-01-1950  |  30-01-1951
  a   |  01-01-1948  |  31-12-1949
  a   |  31-01-1951  |  01-06-2000
  b   |  01-01-1980  |  01-08-2010
  c   |  01-01-1990  |  31-12-2017
  c   |  31-01-1985  |  31-12-1989

我得到了什么

• 每人多行
• 每行一个开始和结束日期
• 按非时间顺序

选择我想返回以下内容的查询:

Select query which I want to return the following:

  ID  |  Date start 1 |  Date end 1  |  Date start 2 |  Date end 2  |  Date start 3 |  Date end 3
---------------------------------------------------------------------------------------------------
  a   |  01-01-1948   |  31-12-1949  |  01-01-1950   |  30-01-1951  |  31-01-1951   |  01-06-2000
  b   |  01-01-1980   |  01-08-2010
  c   |  31-01-1985   |  31-12-1989  |  01-01-1990   |  31-12-2017

我想要的:

• 每人一行
• 每行有多个开始和结束日期
• 按时间顺序

我能找到的大多数东西都希望它在同一列中，或者不希望它按时间顺序排序，所以不幸的是，这些情况不适用于我.

Most things I was able to find wanted it in the same column, or wouldn't want it sorted on chronological order, so unfortunately those situations didn't apply to me.

我现在真的知道如何解决这个问题了.

I really have now clue how to solve this.

推荐答案

如果你只有三个日期，那么pivot/conditional聚合应该没问题:

If you have only three dates, then pivot/conditional aggregation should be fine:

select id,
       max(case when seqnum = 1 then dstart end) as start_1,
       max(case when seqnum = 1 then dend end) as end_1,
       max(case when seqnum = 2 then dstart end) as start_2,
       max(case when seqnum = 2 then dend end) as end_2,
       max(case when seqnum = 3 then dstart end) as start_3,
       max(case when seqnum = 3 then dend end) as end_3
from (select t.*,
             row_number() over (partition by id order by dstart) as seqnum
      from t
     ) t
group by id;

注意:您必须指定输出中的列数.如果您不知道有多少，您可以:

Note: You have to specify the number of columns in the output. If you don't know how many there are, you can either:

• 生成动态 SQL 语句来提前进行计数.
• 手动计数并添加适当的列.

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