CODEWITHSUNDEEP
c#.netwinformstesting
David Elner
David Elner

Reputation: 5181

Form.ShowDialog() does not display window with debugging enabled

I've created a test within a Unit Test Project, in which I want pop up a Form using its ShowDialog() function:

[TestMethod]
public void TestDialog()
{
  // This class inherits from Form
  TestForm serviceTestForm = new TestForm("My test form"); 
  serviceTestForm.ShowDialog();

  return;
}

I expect this test to reach ShowDialog(), and run 'indefinitely', until I close the window. However, when I run this test "with debugging", the test reaches ShowDialog(), and no form appears. Strangely enough, this same exact test works if I run "without debugging."

I need to be able to run the test "with debugging" and have the window display.

Other notes:

Upvotes: 20

Views: 17043

Answers (2)

JotaBe
JotaBe

Reputation: 39004

In my case, using VS2017, setting the property ShowInTaskbar as false did the trick.

This is the complete code used to show the dialog:

form.TopMost = true;
form.StartPosition = FormStartPosition.CenterScreen;
form.ShowInTaskbar = false;
form.ShowDialog();

P.S. After finding this, I've seen the same solution in Displaying Windows Forms inside unit test methods

Upvotes: 0

blins
blins

Reputation: 2535

As much as I try to avoid building unit tests that use System.Windows.Forms, I ran into an odd case where I needed this as well and solved it by handling the Load event and explicitly setting Visible = true. This forces the form to visible when ShowDialog is called from the test method.

private void form1_Load(object sender, EventArgs e)
{
    // To support calling ShowDialog from test method...
    this.Visible = true;
    ...
}

Alternatively, just observe the form instance from your test method and do the same there instead. At least this mitigates the issue further in that it keeps the hack out of your form's code.

var frm = new Form1();
frm.Load += (sender, e) => (sender as Form1).Visible = true;
frm.ShowDialog();

Upvotes: 36

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