在SQL Server中使用数据透视表将列转换为行

Question

我有如下表：

我想将其转换为以下内容：

+------------+---------------------+---------------------+
| Child_Code |     SewingStart     |      SewingEnd      |
+------------+---------------------+---------------------+
|     000001 | 2017-02-21 00:00:00 | 2017-03-21 00:00:00 |
+------------+---------------------+---------------------+

任何帮助请!!

Answer 1

如果行数有限，则可以使用条件聚合或透视。 但是，你需要一个专栏。 所以：

select child_code,
       max(case when seqnum = 1 then plan_date end) as plan_date_1,
       max(case when seqnum = 2 then plan_date end) as plan_date_2
from (select t.*,
             row_number() over (partition by child_code order by plan_date) as seqnum
      from t
     ) t
group by child_code;

如果您知道所需的最大计划日期数，则只能使用此方法。 否则，您将需要使用动态数据透视表。 这个想法是一样的，但是需要构造查询字符串，然后将其传递给sp_executesql 。

编辑：

如果您只有两个值，则group by可能更容易。 以下处理只有一个值的情况：

select child_code, min(plan_date) as plan_date_1,
       (case when min(plan_date) <> max(plan_date) then max(plan_date)
        end) as plan_date_2
from t
group by child_code;

Answer 2

如果plan_date的最大数量未知，则需要使用动态sql。 您需要使用row_number()对Child_Code分区的每个列表进行Child_Code ，以便与pivot() 。

测试设置：

create table t (child_code varchar(6), plan_date datetime);
insert into t values ('000001','20170221'),('000001','20170321');
declare @cols nvarchar(max);
declare @sql  nvarchar(max);

  select @cols = stuff((
    select distinct 
      ',' + quotename('Plan_Date_'
          +convert(nvarchar(10),row_number() over (
              partition by Child_Code 
              order by     Plan_Date 
          ))
          )
      from t 
      for xml path (''), type).value('.','nvarchar(max)')
    ,1,1,'');

select @sql = '
 select Child_Code, ' + @cols + '
 from  (
  select 
      Child_Code
    , Plan_Date
    , rn=''Plan_Date_''+convert(nvarchar(10),row_number() over (
          partition by Child_Code 
          order by     Plan_Date 
        ))
  from t
    ) as a
 pivot (max([Plan_Date]) for [rn] in (' + @cols + ') ) p';
 select @sql as CodeGenerated;
 exec sp_executesql @sql;

rextester 演示 ： http ：//rextester.com/YQCR87525

代码生成：

 select Child_Code, [Plan_Date_1],[Plan_Date_2]
 from  (
  select 
      Child_Code
    , Plan_Date
    , rn='Plan_Date_'+convert(nvarchar(10),row_number() over (
          partition by Child_Code 
          order by     Plan_Date 
        ))
  from t
    ) as a
 pivot (max([Plan_Date]) for [rn] in ([Plan_Date_1],[Plan_Date_2]) ) p

回报

+------------+---------------------+---------------------+
| Child_Code |     Plan_Date_1     |     Plan_Date_2     |
+------------+---------------------+---------------------+
|     000001 | 21.02.2017 00:00:00 | 21.03.2017 00:00:00 |
+------------+---------------------+---------------------+