Hello,
Sorry, i saw this thread a little late. My question is still the same. The solution that you gave actually gives me the display name of the sequence file. But i wanted the name of the sequence that was opened withtin the sequence file. Here is what i have.
A sequence file named TestExport.seq has two sequences - MainSequence and SecondSequence. I open up the sequence file and open the SecondSequence in that file. But i dont know how to get this information about the sequence that is selected within the file. I have an external dll, in which i am passing the sequence file name as RunState.InitialSelection.SelectedFile. But i do not know how to get the sequence name that was actually opened in that file. Because i actually want to get the sequence context of the sequence within the file. The code that i have looks like this:
public ExportDlg(SequenceFile seqFile)
{
InitializeComponent();
SequenceContext seqContext = seqFile.NewEditContext();
engine = seqContext.Engine;
object outParam = System.Type.Missing;
this.seqContext = engine.NewEditContext(
seqFile.GetSequence(0).AsPropertyObject(),System.Type.Missing,out outParam);
}
seqFile.GetSequence(0) - gives me the context of the first sequence which is MainSequence. But if i have opened the SecondSequence, i need the context for that. And to get this, i need the sequence name "SecondSequence" somehow.
RunState.InitialState.SelectedSequences is an array of containers. Although i can see the sequence name during executing, i do not know how to get the name of the container my external dll .
Please help,
Thanks,
Regards,
Aparna