Question

我在SQL服务器上有一个表，我想以xml格式获取数据。

我的表格似乎是

Year | Semester | Room | Professor
-----+----------+------+-----------    
2013 |    1     | 101  | Li
2013 |    1     | 102  | wang
2013 |    1     | 103  | Rob
2013 |    2     | 204  | jess
2013 |    2     | 205  | Mat
2014 |    1     | 104  | Li
2014 |    1     | 105  | wang
2014 |    1     | 106  | Rob
2014 |    2     | 206  | jess
2014 |    2     | 207  | Mat

我想要像这样的Xml格式：

<St>
  <Year id="2013">
    <Semester nb="1">
      <data>
        <Room>101</Room>
        <professor>Li</kwh>
      </data>
      <data>
        <Room>102</Room>
        <professor>wang</kwh>
      </data>
    </Semester>
    <Semester nb="2">
      <data>
        <Room>204</Room>
        <professor>jess</kwh>
      </data>
     </Semester>
  <Year>
 <Year id="2014">
    <Semester nb="1">
     ...
    </Semester>
    <Semester nb="2">
    ...
     </Semester>
  <Year>
</St>

现在我的代码是：

SELECT 
    Year as "@id", 
    Semester as "Semester /@nb", 
    Room as "Semester/data/Room", 
    Professor as "Semester/data/Professor " 
FROM 
    checking
FOR XML Path ('Year'), root ('St')

哪个给我这个xml：

<St>   
<Year id="2013">
<Semester nb="1">
 <data>
        <Room>101</Room>
        <Professor>Li</Professor>
     </data>
   </Semester>   </Year>   <Year id="2013">
    <Semester nb="1">
      <data>
        <Room>102</Room>
        <Professor>wang</Professor>
    </data>
    </Semester>   </Year>   <Year id="2013">
   <Semester nb="1">
     <data>
       <Room>103</Room>
      <Professor>rob</Professor>
     </data>
    </Semester>   
</Year>
 ... 
</St>

我非常接近解决方案，但无法找到如何按年和学期分组。

我尝试分组并按顺序排列但没有一个起作用。

Answer 1

使用分组依据和相关子查询。

declare @T table
(
  Year int,
  Semester int,
  Room int,
  Professor varchar(10)
);

insert into @T values
(2013, 1, 101, 'Li  '),
(2013, 1, 102, 'wang'),
(2013, 1, 103, 'Rob '),
(2013, 2, 204, 'jess'),
(2013, 2, 205, 'Mat '),
(2014, 1, 104, 'Li  '),
(2014, 1, 105, 'wang'),
(2014, 1, 106, 'Rob '),
(2014, 2, 206, 'jess'),
(2014, 2, 207, 'Mat ');

select T1.Year as [@id],
       (
       select T2.Semester as [@nb],
              (
              select T3.Room,
                     T3.Professor
              from @T as T3
              where T3.Year = T2.Year and
                    T3.Semester = T2.Semester
              for xml path('data'), type
              )
       from @T as T2
       where T1.Year = T2.Year
       group by T2.Semester, T2.Year
       for xml path('Semester'), type
       )
from @T as T1
group by T1.Year
for xml path('Year'), root('St');

结果：

<St>
  <Year id="2013">
    <Semester nb="1">
      <data>
        <Room>101</Room>
        <Professor>Li  </Professor>
      </data>
      <data>
        <Room>102</Room>
        <Professor>wang</Professor>
      </data>
      <data>
        <Room>103</Room>
        <Professor>Rob </Professor>
      </data>
    </Semester>
    <Semester nb="2">
      <data>
        <Room>204</Room>
        <Professor>jess</Professor>
      </data>
      <data>
        <Room>205</Room>
        <Professor>Mat </Professor>
      </data>
    </Semester>
  </Year>
  <Year id="2014">
    <Semester nb="1">
      <data>
        <Room>104</Room>
        <Professor>Li  </Professor>
      </data>
      <data>
        <Room>105</Room>
        <Professor>wang</Professor>
      </data>
      <data>
        <Room>106</Room>
        <Professor>Rob </Professor>
      </data>
    </Semester>
    <Semester nb="2">
      <data>
        <Room>206</Room>
        <Professor>jess</Professor>
      </data>
      <data>
        <Room>207</Room>
        <Professor>Mat </Professor>
      </data>
    </Semester>
  </Year>
</St>

Tsql从表创建Xml

1 个答案: