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Filling fields in Word using c# and Microsoft Word Interop

I tried to Fill out Form Fields in Microsoft Word using C# Interop Assemblies with the following Code

string filename = @"N:\mehler\Vorlage2.dotx";

Microsoft.Office.Interop.Word.Application word = new Microsoft.Office.Interop.Word.Application();
Microsoft.Office.Interop.Word.Document doc = new Microsoft.Office.Interop.Word.Document();

doc = word.Documents.Open(filename);
doc.Activate();

foreach (Microsoft.Office.Interop.Word.FormField field in doc.FormFields)
{
    switch (field.Name)
    {
        case "Text2":
            field.Range.Text = "1";
            break;

        default:
            break;
    }
}

doc.SaveAs2(@"N:\mehler\Ausgefuellt.docx");
doc.Close();
word.Quit();
System.Diagnostics.Process.Start(@"N:\mehler\Ausgefuellt.docx");

Microsoft Word can not open the File Ausgefuellt.docx and Shows a Message saying that an unknown Error has occured.

I created a simple Word Document with a bit of unformated text and 2 Text-Form-Fields

can anyone tell me, what went wrong or if ia have an Error in my Source Code

Edit: I managed to specify the Problem. I created an Document only conaining one Text Form Field. In Word 2013 this is found unter the Topic "Formulare aus Vorversionen" (I would translate this to "Formfields from former Versions") If I comment out the whole foreach Block so that I would only open and save the Document, I get the same result.

If I open the Source File directly in Word it is no Problem. I also tried to load the Document and make Word Visible. The Result looked like an empty Word instance with no Document loaded.

like image 647
sebastianmehler Avatar asked Aug 30 '13 12:08

sebastianmehler


2 Answers

You should use:

doc = Word.Documents.Add(filename);

Instead of:

doc = Word.Documents.Open(filename);

So Word will use the template to create a document file, and not open the template itself. It seems Word behaves differently when the active document is a Template.

like image 101
Nir Kornfeld Avatar answered Sep 16 '22 12:09

Nir Kornfeld


Use that, it should work:

Word.Application WordApp;
Word.Document WordDoc;

object misValue = System.Reflection.Missing.Value;

WordApp = new Word.ApplicationClass();
WordDoc = WordApp.Documents.Open(filePath2, misValue, misValue, misValue, misValue, misValue,
        misValue, misValue, misValue, misValue, misValue, misValue, misValue, misValue, misValue);
WordDoc.Activate();
like image 44
Asieh hojatoleslami Avatar answered Sep 17 '22 12:09

Asieh hojatoleslami