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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: 2010-04-05
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
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5 Comments
 
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Accepted Solution

by:
bpana earned 500 total points
ID: 12198715
I think I have already answered a similar question. take a look here:
http://www.experts-exchange.com/Programming/Programming_Languages/Delphi/Q_21046672.html

Bogdan
0
 

Author Comment

by:j_oller
ID: 12201707
Thanks Bogdan,

If my C# web service pushed XML out like the XML on your post, I would not have a problem. 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: 12204985
I got it figured out, I will post the code when I can.

Regards,

Joel
0
 

Author Comment

by:j_oller
ID: 12218673
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:bpana
ID: 12224377
Hi, Joel.

I have used webservices in the same way (exporting data as xml).

Bogdan
0

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