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Example: Spreadsheets from SQL schema

Description

This example creates a multiple spreadsheets from the tables in a SQL::Schema. After the spreadsheet is populated, it is opened in Microsoft Excel.

dim DocumentFile as C = "C:\temp\A5OfficeSample6.xls"
dim ConnectionString as C = "{A5API=Access,FileName='" + a5.get_exe_path() \
                          + "\MDBFiles\Northwind.mdb',UserName='Admin'}"
dim Schema	as SQL::Schema
dim Connection 	as SQL::Connection
 
'We can manipulate the schema any way we need to prior to saving
'Here we are changing the type of the data column for the 
' Category Picture Column and the Employee Photo Column to
' OLEObject (Access doesn't tell us, but we know...)
'ToExcel automatically checks to see if an OLEObject is a 
' bitmap and converts it if it is. If you have binary objects
' that are actually JPG, PNG, DIB, or BMP data, set the value
' accordingly.
if Connection.Open(ConnectionString)
    if Connection.GetSchema(Schema) 
        SetColumnToOLEObject(Schema, "Categories", "Picture")
        SetColumnToOLEObject(Schema, "Employees", "Photo")
        if .not. Schema.ToExcel(DocumentFile,Connection)
            ui_msg_box("Error Saving Schema Tables", \
                       "Error:" + crlf(2) + Schema.callresult.text)	
        else
            sys_open(DocumentFile)
        end if
    else
        ui_msg_box("Error Loading Database Schema", \
                   "Error:" + crlf(2) \
                   + Connection.callresult.text)	
    end if
else
    ui_msg_box("Error:", \
               "Error Connecting To Database" \
               + ConnectionString + crlf(2) + "Error:" \
               + crlf(2) + Connection.callresult.text)	
end if
 
FUNCTION SetColumnToOLEObject as L (Schema as SQL::Schema, \
                                    Table as C, Column as C)
dim Dummy as SQL::IntermediateType 
 
dim TableIndex as N = Schema.TableNumber(Table)
if (TableIndex > 0)
    dim TableInfo as SQL::TableInfo = Schema.Table(TableIndex)
    dim ColumnIndex as N = TableInfo.ColumnNumber(Column)
    if (ColumnIndex > 0)
        TableInfo.Column(ColumnIndex).IntermediateType = OLEObject
        SetColumnToOLEObject = .t.
    end if
end if
END FUNCTION