CODEWITHSUNDEEP
c#xmltemplatesxsdfile-generation
ConstantLearner
ConstantLearner

Reputation: 1

Generating different files from Source in different formats using customized templates

I need to use a source data (basically a database table with different columns and data loaded or a csv file) and then create different files (lets call them destination files) in different formats (text,excel,pdf etc) based on source data using mappings from columns in source data to columns in destination files. The format and no. of columns would be different for different destination files but source data would be same for all. If a new destination file needs to be created in future, the solution design should as simple as just adding another template for the new file and add to existing templates set.

any idea on above approach please? Source data - can be SQL database table or a CSV file. Need to write in C#.NET

Upvotes: -1

Views: 37

Answers (1)

jdweng
jdweng

Reputation: 34421

I like using my CSV reader which takes a CSV file and puts into a DataTable.

    public class CSVReader
    {

        public DataSet ReadCSVFile(string fullPath, bool headerRow)
        {

            string path = fullPath.Substring(0, fullPath.LastIndexOf("\\") + 1);
            string filename = fullPath.Substring(fullPath.LastIndexOf("\\") + 1);
            DataSet ds = new DataSet();

            try
            {
                if (File.Exists(fullPath))
                {
                    string ConStr = string.Format("Provider=Microsoft.Jet.OLEDB.4.0;Data Source={0}" + ";Extended Properties=\"Text;HDR={1};FMT=Delimited\\\"", path, headerRow ? "Yes" : "No");
                    string SQL = string.Format("SELECT * FROM {0}", filename);
                    OleDbDataAdapter adapter = new OleDbDataAdapter(SQL, ConStr);
                    adapter.Fill(ds, "TextFile");
                    ds.Tables[0].TableName = "Table1";
                }
                foreach (DataColumn col in ds.Tables["Table1"].Columns)
                {
                    col.ColumnName = col.ColumnName.Replace(" ", "_");
                }
            }

            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
            return ds;
        }
    }

Upvotes: 0

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