Question

我想知道如何在JavaFX上拍摄节点的多个快照我有这个代码，但不幸的是它没有工作（或者给我想要的东西）

我认为问题出现在线程或循环中

public class FXMLDocumentController implements Initializable {

@FXML
private MediaView media_view;

@FXML
private ImageView image_view;

private Media me;

private MediaPlayer mp;


@Override
public void initialize(URL location, ResourceBundle resources) 
{
    // load a video and play it in the MediaView 
    String s = "/home/mustafa987/Videos/sampels/video10.mp4";
    me = new Media(new File(s).toURI().toString());
    mp = new MediaPlayer(me);
    media_view.setMediaPlayer(mp);
    mp.play();
    mp.setRate(1);
}

//the record() method is invoked when i click a button      
@FXML
void record(ActionEvent event) throws IOException  
{
    Runnable runnable=() -> {
        while (true)
        {
            // snapshot -> buffered image -> image -> display in the ImageView
            WritableImage img = media_view.snapshot(new SnapshotParameters(), null);
            BufferedImage bufImg = SwingFXUtils.fromFXImage(img, null);
            Image image=SwingFXUtils.toFXImage(bufImg, null);
            image_view.setImage(image);
            System.out.println(" - done - ");
            try 
            {
                Thread.sleep(1000); //take snapshot every 1 second 
            } catch (InterruptedException e)
            {
                System.out.println(e);
            }
        }
    };
    Thread thread=new Thread(runnable);
    thread.start();
}
}

Answer 1

由@Pagbo解决来自here

的引用

program test                                                        
    use iso_fortran_env, r8 => real64                               
    implicit none                                                   

    integer, parameter :: ni = 130, nj = 130, nk = 130, nvar = 130  
    real(r8), allocatable :: u1(:,:,:,:), u2(:,:,:,:), w(:,:,:)     
    real(r8) :: sum, t0, t1                                         
    integer :: i,j,k,n                                              

    call cpu_time(t0)                                               
    allocate(u1(ni,nj,nk,nvar))                                     
    allocate(u2(ni,nj,nk,nvar))                                     
    allocate(w(ni,nj,nk))                                           
    call cpu_time(t1)                                               
    write(*,'("allocation time(s):",es15.5)') t1-t0                 

    call cpu_time(t0)                                               
    call random_seed()                                              
    call random_number(u1)                                          
    call random_number(u2)                                          
    call random_number(w)                                           
    call cpu_time(t1)                                               
    write(*,'("random init time (s):",es15.5)') t1-t0               

    sum = 0.0_r8                                                    
    call cpu_time(t0)                                               
    do n = 1, nvar                                                  
        do k = 1, nk                                                
            do j = 1, nj                                            
                do i = 1, ni                                        
                    sum = sum + u1(i,j,k,n)*u2(i,j,k,n)*w(i,j,k)    
                end do                                              
            end do                                                  
        end do                                                      
    end do                                                          
    call cpu_time(t1)                                               
    write(*,'("Sum:",es15.5," time(s):",es15.5)') sum, t1-t0        

end program

如何获得节点JavaFX的多重/连续快照

1 个答案: