Multiple Showdialog

Question

I have a main window which calls another window with ShowDialog(), lets call it window A. Window A calls another window also with ShowDialog(), lets call it window B.

Whenever window B is closed using Close(), I want Window A to be shown, not the main window.

How am I supposed to do that? I already tried using this.Owner, this.Focus, etc. none of them work.

Note: I am using WPF

here is my code from main window:

WindowsA WA = new WindowsA(); WA.showDialog();

at WindowsA, i call another window WindowsB:

WindowsB WB = new WindowsB(); WB.showDialog();

from WindowsB, I pressed button to close:

Close();

Where should I put the owner?

When you do showDialog for window b, shouldn't you set the owner to Window A, Then anything that windows A opens will go back to window a when it's closed. See here: http://stackoverflow.com/a/2045671/4714970 — Aequitas, Jun 22 '15 at 04:07
Can you post some code? When trying to replicate your problem, my dialog B returns to dialog A as you would expect. — Loocid, Jun 22 '15 at 04:09
like @Loocid said when tried to replicate the issue i am also able to return to dialog window A as per your requirement. — Sachu, Jun 22 '15 at 04:13
update the question. I put the simple code. @Aequitas, I already try to set the owner to Window A. But still didn't work. Loocid, Sachu how many form do you have? I 3 forms, main windows, windows A, windows B. — Alfred Angkasa, Jun 22 '15 at 04:21

score 3 · Accepted Answer · answered Jun 29 '15 at 02:03

When you do showDialog for window b, you should set the owner to Window A, Then anything that window A opens will go back to window A when it's closed.

Like this: WA.ShowDialog(this);

This will make this the owner of WA.

See here for more information on this: www.stackoverflow.com/a/2045671/4714970

score 1 · Answer 2 · answered Jun 22 '15 at 05:09

1

Sorry, I just figured it out.

Before WA.ShowDialog(), I just set WA.Owner = this;

After that it works.

Correct me if I'm wrong.

answered Jun 22 '15 at 05:09

Alfred Angkasa

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you said this didn't work before? Also pretty sure you can do `WA.ShowDialog(this);` – Aequitas Jun 22 '15 at 05:54
maybe i forgot to put the 'this' inside the showdialog(). can you just post your answer? i will mark yours as the real answer. thanks – Alfred Angkasa Jun 29 '15 at 01:59

Multiple Showdialog

2 Answers2