Question

我在SQL Server 2014中有一个名为anotes的表，其中包含以下数据

enter image description here

我希望将此数据添加到另一个名为final的表中作为

ID      Notes      NoteDate

text1, text2, text3, text4进入最终表格的Notes列，Notedate1,notedate2,notedate3,notedate4进入Notedate列。

我首先试着用笔记取消数据：

select createdid, temp
from (select createdid,text1,text2,text3,text4 from anotes) p
unpivot
(temp for note in(text1,text2,text3,text4)) as unpvt
order by createdid

这给了我正确的结果：

enter image description here

然后对于日期部分我使用了另一个unpivot查询：

select createdid,temp2
from (select createdid,notedate1,notedate2,notedate3,notedate4 from anotes) p
unpivot (temp2 for notedate in(notedate1,notedate2,notedate3,notedate4)) as unpvt2

这也给了我正确的结果：

enter image description here

现在我想将这些数据添加到我的决赛桌中。

我尝试了以下查询，结果是交叉连接：（

select a.createdid, a.temp, b.temp2
from (select createdid, temp
      from (select createdid,text1,text2,text3,text4 from anotes) p
      unpivot
      (temp for note in(text1,text2,text3,text4)) as unpvt) a inner join (select createdid,temp2
from (select createdid,notedate1,notedate2,notedate3,notedate4 from anotes) p
unpivot (temp2 for notedate in(notedate1,notedate2,notedate3,notedate4)) as unpvt) b on a.createdid=b.createdid

输出如下：

enter image description here

有什么方法可以同时对两个列进行同步？

或者使用两个选择查询将该数据添加到我的最终表中？

提前致谢！

Answer 1

我想说最简洁，也许是最有效的方法来取消多列的删除是使用CROSS APPLY和table valued constructor：

SELECT  t.CreatedID, upvt.Text, upvt.NoteDate
FROM    anotes t
        CROSS APPLY
        (VALUES
            (Text1, NoteDate1),
            (Text2, NoteDate2),
            (Text3, NoteDate3),
            (Text4, NoteDate4),
            (Text5, NoteDate5),
            (Text6, NoteDate6),
            (Text7, NoteDate7)
        ) upvt (Text, NoteDate);

<强> Simplified Example on SQL Fiddle

<强>附录

我觉得这个概念很难解释，但我会试试。值为constuctor的表只是一种动态定义表的方法，所以

SELECT  *
FROM    (VALUES (1, 1), (2, 2)) t (a, b);

将使用带有数据的别名t创建一个表：

a   b
------
1   1
2   2

因此，当您在APPLY中使用它时，您可以访问所有外部列，因此只需使用正确的值对定义构造的表（即text1 with date1）。

Answer 2

使用@AHiggins上面提到的链接

以下是我的最终查询！

select createdid,temp,temp2
from (select createdid,text1,text2,text3,text4,text5,text6,text7,notedate1,notedate2,notedate3,notedate4,notedate5,notedate6,notedate7 from anotes) main
      unpivot
      (temp for notes in(text1,text2,text3,text4,text5,text6,text7)) notes
unpivot (temp2 for notedate in(notedate1,notedate2,notedate3,notedate4,notedate5,notedate6,notedate7)) Dates
where RIGHT(notes,1)=RIGHT(notedate,1)

Answer 3

将每个查询视为一个表，并根据createdid和fieldid（字段名称的数字部分）将它们连接在一起。

select x.createdid, x.textValue, y.dateValue
from
(
    select createdid, substring(note, 5, len(note)) fieldId, textValue
    from (select createdid,text1,text2,text3,text4 from anotes) p
    unpivot
    (textValue for note in(text1,text2,text3,text4)) as unpvt
)x
join
(
    select createdid, substring(notedate, 9, len(notedate)) fieldId, dateValue
    from (select createdid,notedate1,notedate2,notedate3,notedate4 from anotes) p
    unpivot (dateValue for notedate in(notedate1,notedate2,notedate3,notedate4)) as unpvt2
) y on x.fieldId = y.fieldId and x.createdid = y.createdid
order by x.createdid, x.fieldId

如果列数太多且字段名称的最右边数字重复（例如text1和text11），则给出的其他答案将无效。

取消隐藏多列

3 个答案: