我已经将XML文档导入到SQL Server中,现在我正在尝试将各个部分导入不同的表。当我使用以下查询时,它仅返回一排hotel_facities,我需要用hotel_ref返回所有hotel_facities。

DECLARE @Details xml 
    SET @Details = '<hotels>
 <hotel>
  <hotel_ref>105</hotel_ref> 
 <hotel_facilities>
  <id>2</id> 
  <name>Disabled Facilities</name> 
  <id>4</id> 
  <name>24 Hour Reception</name> 
  <id>12</id> 
  <name>Restaurant</name> 
  </hotel_facilities>
  </hotel>
</hotels>'  

SELECT tab.col.value('../hotel_ref[1]','varchar(100)') AS 'hotel_ref',
tab.col.value('./id[1]','varchar(100)') AS 'HotelFacilityID',
tab.col.value('./name[1]','varchar(100)') AS 'HotelFacilityName'
FROM @Details.nodes('//hotels/hotel/hotel_facilities') AS tab(col)
有帮助吗?

解决方案

我同意MARC_S XML没有良好的模式。

我最接近的是:

SELECT tab.col.value('./hotel_ref[1]','varchar(100)') AS 'hotel_ref', 
fac.value('(.)[1]','varchar(100)') AS 'HotelFacilityID', 
ROWID=IDENTITY(int,1,1) 
into #facilitiesid 
FROM @Details.nodes('/hotels/hotel') AS tab(col) 
cross apply col.nodes('.//id') a(fac) 


SELECT tab.col.value('../hotel_ref[1]','varchar(100)') AS 'hotel_ref', 
fac.value('(.)[1]','varchar(100)') AS 'HotelFacilityName', 
ROWID=IDENTITY(int,1,1) 
into #facilitiesnames 
FROM @Details.nodes('//hotels/hotel/hotel_facilities') AS tab(col) 
cross apply col.nodes('.//name') a(fac) 

select i.hotel_ref, HotelFacilityID, HotelFacilityName 
from #facilitiesid i 
inner join #facilitiesnames n 
    on i.rowid = n.rowid 

其他提示

您的XML结构有些有趣 - <hotel_facilities> 不包含适当的“子实体”,您可以列举这……

如果您的设施将被包裹在 <facility>....</facility> 元素,您可以轻松地列举这一点。

   <hotel_facilities>
      <facility>
         <id>2</id> 
         <name>Disabled Facilities</name> 
      </facility>
      <facility>
         <id>4</id> 
         <name>24 Hour Reception</name> 
      </facility>
      <facility>
         <id>12</id> 
         <name>Restaurant</name> 
      </facility>
   </hotel_facilities>

但是,有了您当前的设置,我认为您很难找到一个好的解决方案。

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