2013-03-29 55 views
0
public class WebWiewYahoo extends Application { 
    public static void main(String[] args) { 
     launch(args); 
    } 

    @Override 
    public void start(Stage stage) throws Exception { 
     VBox root = new VBox(); 
     Button go = new Button("go"); 
     final TextField address = new TextField("http://www.yahoo.com"); 
     final WebView browser = new WebView(); 
     final WebEngine webEngine = browser.getEngine(); 
     HBox toolBar1 = new HBox(); 

     toolBar1.setAlignment(Pos.TOP_LEFT); 
     toolBar1.getChildren().addAll(address, go); 

     // load the web page 
     webEngine.load("http://www.yahoo.com"); 

     root.getChildren().add(toolBar1); 
     root.getChildren().add(browser); 
     Scene scene = new Scene(root); 
     stage.setScene(scene); 
     stage.show(); 
    } 
} 

//我想操作瀏覽器中的鏈接。因此,即使點擊任何點的任何鏈接,我可以重定向到yahoo.com.Please幫助。網絡瀏覽器鏈接操作

回答

0

你可以試着這樣做:

webEngine.getLoadWorker().stateProperty().addListener(new ChangeListener<State>() { 
    @Override 
    public void changed(ObservableValue<? extends State> paramObservableValue, 
         State oldState, 
         State newState) { 
     if (State.SCHEDULED == newState && !"http://www.yahoo.com/".equals(webEngine.getLocation())) { 
      webEngine.load("http://www.yahoo.com"); 
     } 
    } 
}); 
+0

我曾嘗試這個代碼,但仍是犯規,即使在這些達到lines..So我不能夠得到國家 –