SQL Server 2005 中的 XML 查询错误答案

【问题标题】：XML query error in SQL Server 2005SQL Server 2005 中的 XML 查询错误
【发布时间】：2011-07-07 13:09:31
【问题描述】：

为了更新大量记录，我使用了 xml 查询。我从前端（C#.net）填充 xml 并将其作为参数传递给存储过程（如 @Master_rows_xml_Update）。

我面临一种情况..当我在存储过程中执行此命令时

declare @i INTEGER,@Master_rows_xml_Update XML;

SET  @Master_rows_xml_Update= '<root>   
                                 <row Id="1" FinYearId="5" EmployeeId="55" EnteredOn="7/1/2011 12:00:00 AM" EnteredBy="1" HouseRentPaid="False" HouseRentAmount="3500.00" officeId="9"/>
                                 <row Id="2" FinYearId="5" EmployeeId="55" EnteredOn="7/1/2011 12:00:00 AM" EnteredBy="1" HouseRentPaid="False" HouseRentAmount="3500.00" officeId="9"/>
                                </root>';

exec sp_xml_preparedocument @i output, @Master_rows_xml_Update;

update EmployeeFinYearInvestment
    set EmployeeFinYearInvestment.FinYearId = ox.FinYearId,
    EmployeeFinYearInvestment.EmployeeId = ox.EmployeeId,
    EmployeeFinYearInvestment.EnteredOn = ox.EnteredOn,
    EmployeeFinYearInvestment.EnteredBy = ox.EnteredBy,
    EmployeeFinYearInvestment.HouseRentPaid = ox.HouseRentPaid,
    EmployeeFinYearInvestment.HouseRentAmount = ox.HouseRentAmount,
    EmployeeFinYearInvestment.officeId=ox.officeId

from OpenXml(@i, 'root/row')
    with (Id INT,FinYearId INT, EmployeeId INT,EnteredOn DATETIME,EnteredBy INT , HouseRentPaid BIT,HouseRentAmount DECIMAL ,officeId INT ) ox
    where EmployeeFinYearInvestment.Id = ox.Id;

exec sp_xml_removedocument @i;

..它无法更新下表的操作...

程序也没有返回任何错误...

任何想法...

谢谢.. 阿尼尔班

【问题讨论】：

您是否尝试调试此代码？它如何选择xml？ EmployeeFinYearInvestment.Id = ox.Id; 子句中的项目是否存在？

标签： sql sql-server-2005

【解决方案1】：

我会尝试摆脱 OpenXml 和那些功能 - 使用 SQL Server 2005 中的新 XQuery/XPath 东西！

试试这个 - 这应该会为您提供 XML 中的值：

DECLARE @XmlUpdate XML 
SET @XmlUpdate  = 
  '<root>   
   <row Id="1" FinYearId="5" EmployeeId="55" EnteredOn="7/1/2011 12:00:00 AM" EnteredBy="1" HouseRentPaid="False" HouseRentAmount="3500.00" officeId="9"/>
   <row Id="2" FinYearId="5" EmployeeId="55" EnteredOn="7/1/2011 12:00:00 AM" EnteredBy="1" HouseRentPaid="False" HouseRentAmount="3500.00" officeId="9"/>
  </root>';

SELECT
    UpdRow.value('(@Id)[1]', 'int') AS 'ID',
    UpdRow.value('(@FinYearId)[1]', 'int') AS 'FinYearID',
    UpdRow.value('(@EmployeeId)[1]', 'int') AS 'EmployeeID',
    UpdRow.value('(@EnteredOn)[1]', 'datetime') AS 'EnteredOn',
    UpdRow.value('(@EnteredBy)[1]', 'int') AS 'EnteredBy',
    UpdRow.value('(@HouseRentPaid)[1]', 'bit') AS 'RentPaid',
    UpdRow.value('(@HouseRentAmount)[1]', 'decimal(20,4)') AS 'RentAmount',
    UpdRow.value('(@OfficeId)[1]', 'int') AS 'OfficeID'
FROM 
    @XmlUpdate.nodes('/root/row') AS R(UpdRow)

当然，如果你可以选择它，你也可以用它来更新！

UPDATE 
    dbo.EmployeeFinYearInvestment
SET
    EmployeeFinYearInvestment.FinYearId = UpdRow.value('(@FinYearId)[1]', 'int'),
    EmployeeFinYearInvestment.EmployeeId = UpdRow.value('(@EmployeeId)[1]', 'int'),
    ..... (and so on). ......
    EmployeeFinYearInvestment.EnteredOn = ox.EnteredOn,
FROM 
    @XmlUpdate.nodes('/root/row') AS R(UpdRow)
WHERE
    EmployeeFinYearInvestment.Id = UpdRow.value('(@Id)[1]', 'int');

【讨论】：