How to select from database table with "SHAPE" column?

Question

I need to select data from a table.

This table has a number of normal columns, like int, varchar, datetime, etc.
However, it also has a "SHAPE" column, which is something custom and special to GIS.

My code throws an exception, saying it can't handle this.

In theory, a SHAPE would just be an array of co-ordinates... so a two-dimensional array of decimals. Perhaps there's some way to convert it to a varchar or string?

Answer 1

What sort of data source are you querying?

What C# libraries are you using to query?

David

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David http://blogs.msdn.com/b/dbrowne/

Answer 2

Hi David,

I'm currently using a self-made class I'm calling "GIS" to access the GIS database.
The data source is Oracle, using Oracle.DataAccess.dll as a project reference.

I run it like so:

DataTable dt = GIS.Run("select * from TABLE");

The Run() function looks like this:

        public static DataTable Run(string str)
        {
            str = str.Replace("@", ":");

            OracleCommand com = new OracleCommand(str, con);
            OracleDataAdapter sda = new OracleDataAdapter(com);
            DataTable dt = new DataTable();
            sda.Fill(dt);

            return dt;
        }

And the connection string:

static OracleConnection con = new OracleConnection("Data Source=(DESCRIPTION =(ADDRESS_LIST =(ADDRESS = (PROTOCOL = TCP)(HOST = " + hostServer + ")(PORT = 1521)))(CONNECT_DATA =(SID = " + SID + ")    )  );User Id=" + username + ";Password=" + password + ";");

Answer 3

So what is the "SHAPE" data type.  AFAIK it's not a built-in type in Oracle.

David

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David http://blogs.msdn.com/b/dbrowne/

Answer 4

Hi CodeSmile,

If it is a User-Defined Type, you need to define a custom type in .NET Framework to mapping it. For detail information, link below is for your reference.

Oracle User-Defined Types (UDTs) and .NET Custom Types

In addition, I suggest you post this question on Oracle Data Provider for .NET official forum for further discussion.

Best Regards,
Li Wang

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