正如@Donnie提到的，听起来像是要做透视，如果是针对SQL Server 2005或更高版本：

with Page (con_Content, cot_Name, pag_Name)
as
(
    select '<p>this could be the left content</p>', 'LeftColumn', 'Tour'
    union
    select '<p>this could be the right content</p>', 'RightColumn', 'Tour'
    union
    select '<p>main content</p>', 'MainContent', 'Tour'
)
select pag_Name, LeftColumn, RightColumn, MainContent
from [Page]
pivot
(
    min(con_Content)
    for cot_Name in (LeftColumn, RightColumn, MainContent)
) as PivotTable
where pag_Name = 'Tour'

如果不是SQL Server 2005+：

/* with cte defined as above */
select pag_Name, 
    max(case cot_Name when 'LeftColumn' then con_Content else '' end) LeftColumn,
    max(case cot_Name when 'RightColumn' then con_Content else '' end) RightColumn,
    max(case cot_Name when 'MainContent' then con_Content else '' end) MainContent
from [Page]
where pag_Name = 'Tour'
group by pag_Name

• • 编辑**

如果透视表列表中的字段没有对应的cot_Name值，查询仍将执行并返回该字段的null。
例如，尝试以下操作：

with Page (con_Content, cot_Name, pag_Name)
as
(
    select '<p>this could be the left content</p>', 'LeftColumn', 'Tour'
    union
    select '<p>main content</p>', 'MainContent', 'Tour'
)
select pag_Name, LeftColumn, RightColumn, MainContent
from [Page]
pivot
(
    min(con_Content)
    for cot_Name in (LeftColumn, RightColumn, MainContent)
) as PivotTable
where pag_Name = 'Tour'

因此，在您的示例中，您可以包含您感兴趣的每个值，并且只需检查null，以查看pag_Name是否包含该cot_Name的任何内容：

/* using cte as defined above */
select pag_Name, LeftColumn, RightColumn, MainContent, MoreContent_1, MoreContent_2 /* etc. */
from [Page]
pivot
(
    min(con_Content)
    for cot_Name in (LeftColumn, RightColumn, MainContent, MoreContent_1, MoreContent_2)
) as PivotTable
where pag_Name = 'Tour'

如果你说每个页面都有左、右和中间的内容，你可以通过在页面表中添加左、右和中间字段来简化它。否则，我个人会在应用程序中处理转换。