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Consuming a C# Web Service’s ADO.NET Dataset in Delphi

Posted on 2004-09-30
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Last Modified: 2008-01-09
I am trying to consume with Delphi an ADO.NET dataset provided by a C# web service. I seem to running into trouble when it comes to having Delphi’s XMLMapper take the XML from the c# web service and map the transformation out properly, it is only returning the column headers, with no data… I am posting the code to the web service.

[WebMethod]
public DataSet GetDS(string sInput)
{
      ds = new DataSet();

      oOracleConn = new OracleConnection();

      sSQL = string.format("select * from table where table.input = {0}", sInput); \\ I simplified this

      oOracleConn.ConnectionString = "User Id=****;Password=********;Data Source=****";

      oDataAdapter = new OracleDataAdapter(sSQL, oOracleConn);

      oOracleConn.Open();
      oDataAdapter.Fill(ds);
      oOracleConn.Close();

      return ds;
}

Please let me know if there are alternatives to this method, along with source code. The end result is for a C# Web Service ADO.NET dataset to be consumed by Delphi 7 and displayed in a data grid. I was not sure where to post this question, so I posted it in C# and Delphi, and whoever can answer my question first, can have points for both posts.

Regards,

Joel
0
Comment
Question by:j_oller
  • 4
  • 4
8 Comments
 
LVL 6

Expert Comment

by:viola123
ID: 12196609
i dont use string.format to format the xml string.
i use xml schema to do that.

eg.
validateXMLString(sInput,validSchemafile);

public void validateXMLString(string sInput, string XMLSchemaFile)
{
// create the XML Reader using the XML string passed
XmlValidatingReader vr = new XmlValidatingReader(sInput, XmlNodeType.Document, null);
                  
// set validation type to schema
vr.ValidationType = ValidationType.Schema;
                  
try
{
// set the schema using the schema in the application settings
vr.Schemas.Add(null, XMLSchemaFile);
}
catch (Exception ex)
{
throw new ApplicationException("Error loading the specified XML Schema. Schema URL:" + XMLSchemaFile + "\r\n" + ex, ex);
}

// set the validation error handler
vr.ValidationEventHandler += new ValidationEventHandler (ValidationHandler);

// initialise class variables for validation handler
xsdValidationCount = 0;
xsdValidationErrorMessage = new StringBuilder(null);

// Ensure the XML string can be validated against the XML schema
while(vr.Read());

// check if any errors where encountered. If so, raise exception
if (xsdValidationCount != 0)
throw new ApplicationException("Error validating XML String.\r\n" + xsdValidationErrorMessage.ToString());
}

/// <summary>
/// Validation error event handler for the XML Validation
/// </summary>
/// <param name="sender"></param>
/// <param name="args"></param>
private void ValidationHandler(object sender, ValidationEventArgs args)
{
++xsdValidationCount;      // increment counter

// build validation message
xsdValidationErrorMessage.Append("Validation error");
xsdValidationErrorMessage.AppendFormat("\tSeverity:{0}", args.Severity);
xsdValidationErrorMessage.AppendFormat("\tMessage:{0}", args.Message);
xsdValidationErrorMessage.Append("\n");
}

try this one to validate your xml string. but you have to creat a schma file for it.

viola
0
 

Author Comment

by:j_oller
ID: 12196762
"i dont use string.format to format the xml string.
i use xml schema to do that." viola

Please show me where I am doing that, to my knowledge I am only employing the string.format to format the SQL statement with the right variable in the select statement, so that part seems to be trivial, since is has nothing to do with the dataset. I am new to SOAP, and web services, so please correct me if I am wrong...

Thanks,

Joel
0
 
LVL 6

Expert Comment

by:viola123
ID: 12197481
string XMLSchemaFile: is a location of the schema file eg."http://localhost/MyWebService/example.xsd"

public void getData(string sInput)
{
    string mysp = yourStoreProcedureName; ///please see the explaination below this method

     oOracleConn = new OracleConnection();
    oOracleConn.ConnectionString = "User Id=****;Password=********;Data   Source=****";


     validateXMLString(sInput,validSchemafile);

     string SELECT_PROC = mysp;

     StringReader objXMLRead = null;
     DataSet objDSXML;
     DataRow objDBRow;
     SqlDataAdapter objAdapter;
     SqlCommand comm;
     DataSet objDSDBTable = null;
   
try
{
// Read XML String into Object
objXMLRead = new StringReader(sInput);
                  
// Create a Dataset and populate with XML
objDSXML = new DataSet();
objDSXML.ReadXml(objXMLRead, XmlReadMode.Auto);

// Create a Dataset for the Table
objDSDBTable = new DataSet("NEWDATASET");

// create a new command
comm = new SqlCommand(SELECT_PROC, oOracleConn);
comm.CommandType = CommandType.StoredProcedure;

// Create Adapter and fill the dataset
objAdapter = new SqlDataAdapter(comm);

objAdapter.Fill(objDSDBTable);

// loop through XML dataset
foreach (DataRow objDataRow in objDSXML.Tables[0].Rows)
{
     // loop through columns and set new data
      foreach (DataColumn objCol in objDSXML.Tables[0].Columns)
     {
         if (objDataRow[objCol.ColumnName] != DBNull.Value)
         {
           objDBRow[objCol.ColumnName] = objDataRow[objCol.ColumnName];
         }
     }

    // add the new row
     objDSDBTable.Tables[0].Rows.Add(objDBRow);            
}
}
catch (Exception ex)
{
   ........
}

}
***********************************
create procedure yourStoreProcedureName
as
 select a1, a2, ........ from Table where 1=2

go
******************
you can modify this method to suit yours.

viola
0
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Author Comment

by:j_oller
ID: 12201760
Viola, thanks for the quick respone, however, the problem seems to be with the XML that my Web Service is returning as a dataset. I will paste a little more code to illustrate my problem.

C# Webservice:

          [WebMethod]
          public DataSet GetDS2()
          {
         DataTable dt = new DataTable();
          dt.Columns.Add("ID", System.Type.GetType("System.Int32"));
         dt.Columns.Add("Data", System.Type.GetType("System.String"));
          dt.Columns.Add("CreateDate", System.Type.GetType("System.DateTime"));

          DataRow dr;

          dr = dt.NewRow();
          dr["ID"] = 1;
          dr["Data"] = "First Row";
          dr["CreateDate"] = System.DateTime.Now;
         dt.Rows.Add(dr);

          dr = dt.NewRow();
          dr["ID"] = 2;
          dr["Data"] = "Second Row";
          dr["CreateDate"] = System.DateTime.Now;
          dt.Rows.Add(dr);

          DataSet ds2 = new DataSet("SampleDataSet");
          ds2.Tables.Add(dt);
          return ds2;
          }

Pushes this XML, which Delphi's XML Mapper will not map with dataset:

<?xml version="1.0" encoding="utf-8"?>
<DataSet xmlns="http://tempuri.org/">
  <xs:schema id="SampleDataSet" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
    <xs:element name="SampleDataSet" msdata:IsDataSet="true">
      <xs:complexType>
        <xs:choice maxOccurs="unbounded">
          <xs:element name="Table1">
            <xs:complexType>
              <xs:sequence>
                <xs:element name="ID" type="xs:int" minOccurs="0" />
                <xs:element name="Data" type="xs:string" minOccurs="0" />
                <xs:element name="CreateDate" type="xs:dateTime" minOccurs="0" />
              </xs:sequence>
            </xs:complexType>
          </xs:element>
        </xs:choice>
      </xs:complexType>
    </xs:element>
  </xs:schema>
  <diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
    <SampleDataSet xmlns="">
      <Table1 diffgr:id="Table11" msdata:rowOrder="0" diffgr:hasChanges="inserted">
        <ID>1</ID>
        <Data>First Row</Data>
        <CreateDate>2004-10-01T09:07:03.9186134-07:00</CreateDate>
      </Table1>
      <Table1 diffgr:id="Table12" msdata:rowOrder="1" diffgr:hasChanges="inserted">
        <ID>2</ID>
        <Data>Second Row</Data>
        <CreateDate>2004-10-01T09:07:03.9186134-07:00</CreateDate>
      </Table1>
    </SampleDataSet>
  </diffgr:diffgram>
</DataSet>

The XML needs to look like this instead (I got this from C# Windows Application -> ds2.WriteXml("c:\\output.xml")):

<?xml version="1.0" standalone="yes"?>
<SampleDataSet>
  <Table1>
    <ID>1</ID>
    <Data>First Row</Data>
    <CreateDate>2004-10-01T09:15:12.1248693-07:00</CreateDate>
  </Table1>
  <Table1>
    <ID>2</ID>
    <Data>Second Row</Data>
    <CreateDate>2004-10-01T09:15:12.1248693-07:00</CreateDate>
  </Table1>
</SampleDataSet>

Thanks,

Joel
0
 

Author Comment

by:j_oller
ID: 12204983
I got it figured out, I will post the code when I can.

Regards,

Joel
0
 
LVL 6

Accepted Solution

by:
viola123 earned 500 total points
ID: 12213863
hi, paste your code here. so i can see what's going on.

viola
0
 

Author Comment

by:j_oller
ID: 12218626
Before: I was returning a Dataset document, which most other applications dont know how to handle
Now: I have changed my code to return a plain XML Document, and then converted my dataset to xml, and served it through the webservice

There may be better ways of doing it, but this is what I have come up with.

[WebMethod]
            public System.Xml.XmlDocument getDS2()
            {
                  DataTable dt = new DataTable();
                  DataSet ds2 = new DataSet("SampleDataSet");

                  dt.Columns.Add("ID", System.Type.GetType("System.Int32"));
                  dt.Columns.Add("Data", System.Type.GetType("System.String"));
                  dt.Columns.Add("CreateDate", System.Type.GetType("System.DateTime"));

                  DataRow dr;

                  dr = dt.NewRow();
                  dr["ID"] = 1;
                  dr["Data"] = "First Row";
                  dr["CreateDate"] = System.DateTime.Now;
                  dt.Rows.Add(dr);

                  dr = dt.NewRow();
                  dr["ID"] = 2;
                  dr["Data"] = "Second Row";
                  dr["CreateDate"] = System.DateTime.Now;
                  dt.Rows.Add(dr);

                  ds2.Tables.Add(dt);
            
                  System.IO.StringWriter sw = new System.IO.StringWriter();
                  ds2.Namespace="";
                  ds2.WriteXml(sw);
                  System.Xml.XmlDocument xDoc = new System.Xml.XmlDocument();
                  xDoc.LoadXml(sw.ToString());
                  return xDoc;
            }

Thanks,

Joel
0
 
LVL 6

Expert Comment

by:viola123
ID: 12222294
thank you, joel.

viola
0

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