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我正在尝试创建数据源的 XElement 以插入 SSRS RDL 文件。但是,对于 rd: 别名,我似乎无法正确创建它。这是我正在使用的代码。

XNamespace rootNs = "http://schemas.microsoft.com/sqlserver/reporting/2008/01/reportdefinition";
XNamespace rdNs = "http://schemas.microsoft.com/SQLServer/reporting/reportdesigner";
XElement _dataSource = new XElement("DataSource",                            
             new XAttribute(XNamespace.Xmlns + "rd", rdNs),
             new XAttribute("Name", "eFinancials_LOCAL"),
             new XElement("ConnectionProperties", new XElement("DataProvider", "SQL"), new XElement("ConnectString", connectionString)),
             new XElement(_rdns + "SecurityType", "DataBase"),
             new XElement(_rdns + "DataSourceID", dataSourceId)
                                            );

生成的 XML 元素是:

<DataSource xmlns:rd="http://schemas.microsoft.com/SQLServer/reporting/reportdesigner" Name="eFinancials_LOCAL">
  <ConnectionProperties>
    <DataProvider>SQL</DataProvider>
    <ConnectString>Data Source=.;Initial Catalog=800_LMS_eFin_Deploy</ConnectString>
  </ConnectionProperties>
  <SecurityType xmlns="rd">DataBase</SecurityType>
  <DataSourceID xmlns="rd">56e5e869-6ca5-44f9-8340-22821177569e</DataSourceID>
</DataSource>

但是,它必须是:

<DataSource Name="eFinancials_LOCAL">
  <ConnectionProperties>
    <DataProvider>SQL</DataProvider>
    <ConnectString>Data Source=.;Initial Catalog=800_LMS_eFin_Deploy</ConnectString>
  </ConnectionProperties>
  <rd:SecurityType>DataBase</SecurityType>
  <rd:DataSourceID>56e5e869-6ca5-44f9-8340-22821177569e</DataSourceID>
</DataSource>

如何调整我的代码以创建上述正确的 XML?我几乎到了将其创建为文本的地步。

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1 回答 1

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XNamespace ns = @"http://www.somesite.com/xml/customer/2006-10-31";
XElement xe = new XElement(ns + "customers", 
   new XAttribute("xmlns", ns),
   new XElement(ns + "customer", new XElement(ns + "firstname", customer.FirstName); 
于 2013-05-03T19:43:26.173 回答