I am testing this EDI standard: X12 Parser (link), now example in link have as result.txt. The code that does this is:
using OopFactory.X12.Parsing;
using OopFactory.X12.Parsing.Model;
namespace MyX12.Edi835Parser
{
class Program
{
static void Main(string[] args)
{
Stream transformStream = Assembly.GetExecutingAssembly().GetManifestResourceStream("MyX12.Edi835Parser.X12-835-To-CSV.xslt");
Stream inputStream = new FileStream(args[0], FileMode.Open, FileAccess.Read);
Stream outputFile = new FileStream(args[1], FileMode.Create, FileAccess.Write);
X12Parser parser = new X12Parser();
Interchange interchange = parser.Parse(inputStream);
string xml = interchange.Serialize();
var transform = new XslCompiledTransform();
transform.Load(XmlReader.Create(transformStream));
transform.Transform(XmlReader.Create(new StringReader(xml)), new XsltArgumentList(), outputFile);
}
}
}
As you can see, the code has: Stream outputFile = new FileStream(args1 ... where args1 is in project properties / Debug set as Sample-Output.txt, which is the name of file that would be created.
Now, I want to have the result instead as Sample-Output.txt, in my console, something like this:
Stream outputFile = Console.Write();
Really thanks for help.
Console.OpenStandardOutput()
acquires the standard output stream.
Try replacing
Stream outputFile = new FileStream(args[1], FileMode.Create, FileAccess.Write);
with
Stream outputFile = Console.OpenStandardOutput();