Good day, I need some help, I have a dataset which I populate from my db, I also have a xsd Schema file I need to create an xml file using the data from the db and the xsd file.
Can anyone please help me.
All the best
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Use following : DataSet ds = new DataSet(); ds.WriteXml("Filename", XmlWriteMode.WriteSchema);jdweng– jdweng2015-07-01 10:04:24 +00:00Commented Jul 1, 2015 at 10:04
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I already have the schema file, i need to basically combine my db data and my schema to create the xml fileStefan Jacobs– Stefan Jacobs2015-07-01 13:48:56 +00:00Commented Jul 1, 2015 at 13:48
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You don't need to write the schema, but I recommend comparing you existing schema against the generated schema. If the two schemas are the same, then just used the generated one. Makes code a lot simpler.jdweng– jdweng2015-07-01 17:18:27 +00:00Commented Jul 1, 2015 at 17:18
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Do you possibly have a code example for me as I am very new to xml. Do my db column names have to be the same as my xml attriute names?Stefan Jacobs– Stefan Jacobs2015-07-02 07:58:27 +00:00Commented Jul 2, 2015 at 7:58
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1 Answer
You can cross reference the database names with the datatable names in you Select Query by using "as" : "Select abc as xyz from table1". "abc" is the database name and xyz is the datatable name which will be the xml tag name.
Here is an example of writing a DataSet
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.IO;
using System.Data;
namespace ConsoleApplication34
{
class Program
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
DataTable dt = new DataTable("MyTable");
dt.Columns.Add("Name", typeof(string));
dt.Columns.Add("Id", typeof(int));
dt.Rows.Add(new object[] { "John", 1});
dt.Rows.Add(new object[] { "Mary", 2});
dt.Rows.Add(new object[] { "Dick", 3});
dt.Rows.Add(new object[] { "Harry", 4});
dt.Rows.Add(new object[] { "Jane", 5});
DataSet ds = new DataSet("MySet");
ds.Tables.Add(dt);
ds.WriteXml(FILENAME, XmlWriteMode.WriteSchema);
// or
//ds.WriteXml(FILENAME);
}
}
}
