Question

我已经创建了WPF应用。关闭窗口时（通过单击标题栏上的X），我正在隐藏应用程序，我们可以从任务栏图标重新启动或退出该应用程序。

       function updatedata(Request $req)
   {

     dd($req->input());

     $post=User::find($req->input('userid'));
     $post->name=$req->input('name');
        $post->address=$req->input('address');
        $post->photo=$req->input('image');
        $post->gender=$req->input('gender');
        $post->hobbies= implode(',', Input::get('hobbies'));
        $post->std=$req->input('std');
        $requestData = $req->all();
       if($req->hasFile('image'))
       {
        $image=$req->file('image');
        $rules = array('file' => 'required|mimes:png,gif,jpeg'); // 'required|mimes:png,gif,jpeg,txt,pdf,doc'
        $validator = \Illuminate\Support\Facades\Validator::make(array('file'=> $file), $rules);
        $pathToStore=public_path('/');

        if($validator->passes()) 
        {
            $filename = $file->getClientOriginalName(); 
            $extension = $file -> getClientOriginalExtension();
            $picture = sha1($filename . time()) . '.' . $extension;
            $upload_success = $file->move($pathToStore, $picture);

            if($upload_success)
            {
                //if success, create thumb
                $image = Image::make($picture)->resize(600, 531)->save($pathToStore,$picture);
                $oldfilename=$user->photo;
                $user->photo=$filename;
                Storage::delete($oldfilename);  
            }
        }
        $requestData['photo'] = "$pathToStore/{$picture}";     
      }
      $post->update($requestData);
      return redirect()->route('show');     
   }

通知图标对我而言很好，即当我双击任务栏图标App即将降落。以下是双击事件的代码

        NotifyIcon notifyIcon = new NotifyIcon();
        string a = Properties.Resources.SuccessTitle;
        notifyIcon.Icon = Properties.Resources.pencil;
        notifyIcon.ContextMenu = new ContextMenu(new MenuItem[]
            {  exitMenuItem });
        notifyIcon.Visible = true;
        notifyIcon.Text = "My App Name";
        notifyIcon.DoubleClick += notifyIcon_DoubleClick;

当应用程序处于最大化或最小化状态时，单击exe也可以正常工作。

if (window.IsVisible)
        {
            if (window.WindowState == WindowState.Minimized)
            {
                window.WindowState = WindowState.Normal;
                window.Activate();
            }
            else
            {
                window.Activate();
            }
        }
        else
        {
            window.Visibility = Visibility.Visible;
        }

现在，在隐藏窗口的情况下单击exe时，我希望具有相同的效果。我该如何实现。我尝试调试代码，发现当应用隐藏时

MainWindowHandle = 0X00000000

，并且在可见或最小化或最大化时，MainWindowHandle具有一些有效值。

将wpf应用程序带到在后台运行的Foreground？

0 个答案: