Autodetect proxy - JavaFX - webview - webview

My browser (webview) starts with an HTML page
FILEJAVA.class.getResource ("FILEHTML.html"). ToExternalForm ()
Whenever I access the google, I want to know whether the browser check, if the network has proxy (proxy'm working manual)
So that the browser shows a dialog to enter User name and password.

You can use ProxySelector to check proxy. See next example:
public class DetectProxy extends Application {
private Pane root;
#Override
public void start(final Stage stage) throws URISyntaxException {
root = new VBox();
List<Proxy> proxies = ProxySelector.getDefault().select(new URI("http://google.com"));
final Proxy proxy = proxies.get(0); // ignoring multiple proxies to simplify code snippet
if (proxy.type() != Proxy.Type.DIRECT) {
// you can change that to dialog using separate Stage
final TextField login = new TextField("login");
final PasswordField pwd = new PasswordField();
Button btn = new Button("Submit");
btn.setOnAction(new EventHandler<ActionEvent>() {
#Override
public void handle(ActionEvent t) {
System.setProperty("http.proxyUser", login.getText());
System.setProperty("http.proxyPassword", pwd.getText());
showWebView();
}
});
root.getChildren().addAll(login, pwd, btn);
} else {
showWebView();
}
stage.setScene(new Scene(root, 600, 600));
stage.show();
}
private void showWebView() {
root.getChildren().clear();
WebView webView = new WebView();
final WebEngine webEngine = webView.getEngine();
root.getChildren().addAll(webView);
webEngine.load("http://google.com");
}
public static void main(String[] args) {
launch();
}
}
authentification may require additional code in some cases, see Authenticated HTTP proxy with Java for details.

Related

Vaadin Grid middle mouse click

I'm trying to emulate normal browser behaviour in my vaadin grid, which includes middle mouse click to open in a new tab:
addItemClickListener(e -> {
boolean newTab = e.getMouseEventDetails().getButton() == MouseEventDetails.MouseButton.MIDDLE || e.getMouseEventDetails().isCtrlKey();
//open in window or new tab
});
However, the middle mouse button is not registered by vaadin. How could I get this to work?
That feature was included in vaadin-grid (which goes into Vaadin 10) and will not work in Vaadin 8.
For Vaadin 8, you can either intercept the event with some client-side extension, or use a ComponentRenderer for adding a Panel to each component (which works, but is not ideal because it degrades performance):
grid.addColumn(item->{
Panel p = new Panel(item.getName());
p.setStyleName(ValoTheme.PANEL_BORDERLESS);
p.addClickListener(ev->{
System.out.println(ev.getButtonName());
});
return p;
}).setRenderer(new ComponentRenderer());
A client-side extension, on the other hand, allows listening to javascript events (such as MouseEvent) and triggering a server event in response. Creating a extension is quite a complex topic (since it uses a part of the API that is normally hidden from the developer) but it allows direct access to rendered DOM, which is not possible otherwise.
The following resources from the documentation may give you a starting point:
Creating a component extension (which describes a simple extension with Java code only) and Integrating JavaScript Components and Extension (which explains how to add native JavaScript code to your extension).
How I solved the problem in my specific case:
Server side:
public class MyGrid<T> extends Grid<T> {
public MyGrid(String caption, DataProvider<T, ?> dataProvider) {
super(caption, dataProvider);
MiddleClickExtension.extend(this);
}
public static class MiddleClickExtension<T> extends AbstractGridExtension<T> {
private MiddleClickExtension(MyGrid<T> grid) {
super.extend(grid);
registerRpc((rowKey, columnInternalId, details) -> grid.fireEvent(
new ItemClick<>(grid, grid.getColumnByInternalId(columnInternalId), grid.getDataCommunicator().getKeyMapper().get(rowKey), details)),
MiddleClickGridExtensionConnector.Rpc.class);
}
public static void extend(MyGrid<?> grid) {
new MiddleClickExtension<>(grid);
}
#Override
public void generateData(Object item, JsonObject jsonObject) {
}
#Override
public void destroyData(Object item) {
}
#Override
public void destroyAllData() {
}
#Override
public void refreshData(Object item) {
}
}
}
Client side:
#Connect(MyGrid.MiddleClickExtension.class)
public class MiddleClickGridExtensionConnector extends AbstractExtensionConnector {
#Override
protected void extend(ServerConnector target) {
getParent().getWidget().addDomHandler(event -> {
if (event.getNativeButton() == NativeEvent.BUTTON_MIDDLE) {
event.preventDefault();
CellReference<JsonObject> cell = getParent().getWidget().getEventCell();
getRpcProxy(Rpc.class).middleClick(cell.getRow().getString(DataCommunicatorConstants.KEY), getParent().getColumnId(cell.getColumn()),
MouseEventDetailsBuilder.buildMouseEventDetails(event.getNativeEvent(), event.getRelativeElement()));
}
}, MouseDownEvent.getType());
}
#Override
public GridConnector getParent() {
return (GridConnector) super.getParent();
}
public interface Rpc extends ServerRpc {
void middleClick(String rowKey, String columnInternalId, MouseEventDetails details);
}
}

JavaFX WebView cookie

How do I get the cookie set by a connection to a webpage from JavaFX WebView. I want to use this cookie in order to open a seperate connection to the website after the originaol login.
Is there a way to do this and how?
Thank you for your time
I have a local "login.html" that does an AJAX call to my server to log in.
JavaFX application code:
public class Main extends Application {
public static void main(String[] args) {
launch(args);
}
#Override
public void start(Stage primaryStage) {
primaryStage.setTitle("Battleround client");
URL loginPageUrl = Main.class.getResource("/pages/login.html");
final WebView webview = new WebView();
final WebEngine webEngine = webview.getEngine();
webEngine.setJavaScriptEnabled(true);
webEngine.load(loginPageUrl.toString());
/*
* Alright, this piece of code might be hard to understand. Basically
* we're adding a JavaScript object that's actually a Java object. So we
* can call Java methods from JavaScript. And we're adding this
* javascript object as soon as the page has been fully loaded.
*/
webEngine.getLoadWorker().stateProperty().addListener(new ChangeListener<Worker.State>() {
#Override
public void changed(ObservableValue<? extends State> ov, State t, State t1) {
if (t1 == Worker.State.SUCCEEDED) {
JSObject window = (JSObject) webEngine.executeScript("window");
window.setMember("java", new AuthenticationApplication());
}
}
});
StackPane root = new StackPane();
root.getChildren().add(webview);
primaryStage.setScene(new Scene(root, 400, 500));
primaryStage.show();
}
public class AuthenticationApplication {
public void start(String JSESSIONID) {
// From here on I start my game with the JSESSIONID from the login call.
}
}
My JavaScript code in the login.html:
function login(username, password) {
var xmlHttp = new XMLHttpRequest();
if (window.ActiveXObject) {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var url = "http://192.168.2.7:8888/Login";
var params = "username=" + userName + "&password=" + password;
xmlHttp.open("POST", url, true);
// Send the proper header information along with the request
xmlHttp.setRequestHeader("Content-type",
"application/x-www-form-urlencoded");
xmlHttp.setRequestHeader("Content-length", params.length);
xmlHttp.setRequestHeader("Connection", "close");
xmlHttp.withCredentials = "true";
xmlHttp.onreadystatechange = function() {// Call a function when the state changes.
if (xmlHttp.readyState == 4 && xmlHttp.status == 200) {
if(xmlHttp.responseText == "login.successful") {
document.getElementById("loginForm").innerHTML = getResources("login.successful");
var setCookieHeader = xmlHttp.getResponseHeader('Set-Cookie');
java.start(setCookieHeader.split(";")[0].split("=")[1]); // Obtain the JSESSIONID, and send it to the java code. Warning: this code will mess up if there's other cookies.
} else {
document.getElementById("validationMessage").innerHTML = getResources(xmlHttp.responseText);
}
}
};
xmlHttp.send(params);
}
function getResources(key) {
var resources = {};
resources["login.already.logged.in"] = "You are already logged in! Log out first if you want to try again.";
resources["login.error"] = "The login failed because an error occurred, Sorry!";
resources["login.failed"] = "User name and password do not match.";
resources["login.successful"] = "Login succeeded.";
return resources[key];
}

Launching the application from an url in the browser for BlackBerry?

I am developing one application where i will launch a url in the browser from which i will launch my application.
Suppose if i will click google.com, and press enter, it will launch my application. For that i tried with the HttpFilterRegistry API.
For reference i am using the HTTPFilterDemo application. But currently while launching the app, i am getting the NullPointerException.
I wrote the below code i the openFilter Method:
public Connection openFilter(String name, int mode, boolean timeouts) throws IOException {
Logger.out("Protocol", "it is inside the openFilter method");
_url = name.substring(2);
_requestHeaders = new HttpHeaders();
_responseHeaders = new HttpHeaders();
_responseHeaders.setProperty(HttpProtocolConstants.HEADER_CONTENT_TYPE, "text/html");
Logger.out("Protocol", "here it is come ::::44444444");
final int modHandle = CodeModuleManager.getModuleHandle("AppLaunchBrowser");
Logger.out("Protocol", "here is the module handle:::" + modHandle);
final ApplicationDescriptor[] apDes = CodeModuleManager.getApplicationDescriptors(modHandle);
final ApplicationDescriptor appDescriptor = new ApplicationDescriptor(apDes[0], new String[] {});
Logger.out("Protocol", "here is the app descriptor:::" + appDescriptor);
try {
final int appCode = ApplicationManager.getApplicationManager().runApplication(appDescriptor, true);
Logger.out("Protocol", "here is the app code:::" + appCode);
} catch (ApplicationManagerException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
// }
return this;
}
And in the application class i am creating alternative entry point and using like below:
public class AppLaunch extends UiApplication{
public static void main(String args[])
{
Logger.out("AppLaunch", args+"length of the arguments::::" +args.length);
if((args != null) && (args.length > 0) && (args[0].equals("background")))
{
Logger.out("AppLaunch", "in the alternate entry point");
// Logger.out("AppLaunch", args+"length of the arguments::::" +args.length);
HttpFilterRegistry.registerFilter("www.google.co.in", "com.innominds.ca", false);
}
else
{
Logger.out("AppLaunch", "Inside the Applaunch");
AppLaunch theApp = new AppLaunch();
theApp.requestForeground();
Logger.out("AppLaunch", "created the app launch object");
theApp.enterEventDispatcher();
// Logger.out("AppLaunch", "in the alternate entry point");
// HttpFilterRegistry.registerFilter("www.google.co.in", "com.innominds.ca", false);
}
}
public AppLaunch()
{
checkPermissions();
showTestScreen();
}
private void checkPermissions()
{
ApplicationPermissionsManager apm = ApplicationPermissionsManager.getInstance();
ApplicationPermissions original = apm.getApplicationPermissions();
if(original.getPermission(ApplicationPermissions.PERMISSION_BROWSER_FILTER) == ApplicationPermissions.VALUE_ALLOW)
{
// All of the necessary permissions are currently available
return;
}
ApplicationPermissions permRequest = new ApplicationPermissions();
permRequest.addPermission(ApplicationPermissions.PERMISSION_BROWSER_FILTER);
boolean acceptance = ApplicationPermissionsManager.getInstance().invokePermissionsRequest(permRequest);
if(acceptance)
{
// User has accepted all of the permissions
return;
}
else
{
}
}
private void showTestScreen()
{
UiApplication.getUiApplication().pushScreen(new AppLaunchScreen());
}
}
Finally i was able to resolve this issue. Actually NPE is coming in some other callback methods because i was implementing the FilterBaseInterface.

previous instance still active error in blackberry

I created app which user can start from menu and from icon. I do not use GlobalEventListener in my app, just register ApplicationMenuitem. And now I am getting error: previous instance still active when launch my app.
Steps to reproduce not so trivial:
launch app from icon
do not close it, just switch to another app
launch app from icon again
I founded article in blackberry's forum about it , but I can't find solution where I should remove my ApplicationMenuItem: it added on phone boot and should show all the time.
My code:
public class Jingu extends UiApplication {
public static void main(String[] args) {
ApplicationManager app = ApplicationManager.getApplicationManager();
boolean keepGoing = true;
while (keepGoing) {
if (app.inStartup()) {
try {
Thread.sleep(1000);
} catch (Exception e) {}
} else {
keepGoing = false;
}
}
Jingu theApp = new Jingu();
theApp.initMenuItem();
theApp.showMainScreen();
theApp.enterEventDispatcher();
}
public Jingu() {
}
public void showMainScreen() {
showScreen(new JinguMainScreen(this));
}
public void initMenuItem() {
// Create menu item
Object o = RuntimeStore.getRuntimeStore().get(JinguMenuItem.MY_MENU_ID);
// register only if not done already.
if (o == null) {
new JinguMenuItem(this).registerInstance();
}
}
public void showScreen(Screen aScreen) {
synchronized (Application.getEventLock()) {
try {
UiApplication.getUiApplication().popScreen(aScreen);
} catch (Exception e) {
}
UiApplication.getUiApplication().pushScreen(aScreen);
}
}
}
public class JinguMenuItem extends ApplicationMenuItem {
public static final long MY_MENU_ID = 0xb9739d5240d5943dL;
private final Jingu jingu;
public JinguMenuItem(Jingu jingu) {
super(0x350100);
this.jingu = jingu;
}
public void registerInstance() {
Object menuItem = RuntimeStore.getRuntimeStore().remove(MY_MENU_ID);
if (menuItem == null) {
ApplicationMenuItemRepository amir = ApplicationMenuItemRepository.getInstance();
amir.addMenuItem(ApplicationMenuItemRepository.MENUITEM_SYSTEM, this);
RuntimeStore.getRuntimeStore().put(MY_MENU_ID, this);
}
}
public Object run(Object context) {
jingu.setDefaultFont(Font.getDefault());
jingu.setMainApp(false);
jingu.setBbmEditField(null);
jingu.showMainScreen();
return context;
}
public String toString() {
return "My Menu";
}
}
plz advice where I should delete ApplicationMenuItem in my app?
my regards,
Vadim
If you are registering an ApplicationMenuItem from your application, as a user I would consider it bad style for your application to remove and exit, even if RIM provided a way to do this. You may want to separate your application into two parts. One provides the minimal support for responding to the ApplicationMenuItem selection, that starts automatically and runs in the background. The other has all the rest and can run and exit as needed.
My solution for this situation is:
create alternative entry point and run it on app load
register menu in it
do not use runtimeStore

Twitter Follow Link

How do I create a link that will automatically make a user follow a certain Twitter user if they're logged in or send them to Twitter to login first if they're not? I had found how to do this about month or 2 ago but can't find it again. I think it was something basic like a link or a form post to something like twitter.com/[user]/follow.
I've looked at the API, but I'd need the user to authenticate themselves on my site, and I don't want to deal with that. I just want them to authenticate directly on Twitter and not worry about it. The way I had found was nice and simple and I just want to find that again.
Use Twitter's web intents.
While you can use the follow button, you can also send users directly to the Intent URL, like so:
https://twitter.com/intent/user?screen_name=NASA
how to use twitter api in my android application to implement follow button only
Android
http://code.google.com/p/android-hackathon-in-fukuoka/source/browse/trunk/sodefuri/src/jp/jagfukuoka/sodefuri/TimeLineActivity.java?spec=svn167&r=167
Code Snip: (I have converted chines string into standard English)
public class TimeLineActivity extends ListActivity {
private TwitterPreferenceManager tpm = new TwitterPreferenceManager(this);
private static final int FOLLOW = 1;
private static final CharSequence FOLLOW_LABEL = "Follow";
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
// timeline Obtaining process
String screenName = getIntent().getStringExtra("screen_name");
List<String> list = this.getTimeLine(screenName);
setListAdapter(new ArrayAdapter<String>(this, R.layout.timeline_item,list));
}
#Override
public boolean onCreateOptionsMenu(Menu menu) {
menu.add(0, FOLLOW, 0, FOLLOW_LABEL);
return super.onCreateOptionsMenu(menu);
}
#Override
public boolean onOptionsItemSelected(MenuItem item) {
switch (item.getItemId()) {
case FOLLOW:
ConfigurationBuilder builder = new ConfigurationBuilder();
Configuration conf = builder.setOAuthAccessToken(tpm.getAccessToken())
.setOAuthAccessTokenSecret(tpm.getAccessTokenSercret())
.setOAuthConsumerKey(TwitterPreferenceManager.CONSUMER_KEY)
.setOAuthConsumerSecret(TwitterPreferenceManager.CONSUMER_SERCRET)
.setDebugEnabled(true)
.build();
Twitter twitter = new TwitterFactory(conf).getInstance();
try {
String screen_name = getIntent().getStringExtra("screen_name");
twitter.createFriendship(screen_name);
Toast.makeText(getApplicationContext(), "Was to follow.", Toast.LENGTH_LONG).show();
} catch (TwitterException e) {
Toast.makeText(getApplicationContext(), e.getMessage(), Toast.LENGTH_LONG).show();
e.printStackTrace();
}
break;
default:
break;
}
return super.onOptionsItemSelected(item);
}
/**
* Get the time line for the specified user
*
* #param screenName
* #return
*/
private List<String> getTimeLine(String screenName) {
List<String> result = new ArrayList<String>();
Twitter twitter = new TwitterFactory().getInstance();
ResponseList<Status> userTimeline;
try {
userTimeline = twitter.getUserTimeline(screenName);
for (Status status : userTimeline) {
result.add(status.getText());
}
} catch (TwitterException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return result;
}
}
iPhone
http://www.chrismaddern.com/twitter-follow-button-for-ios-iphone-code/
Here is the way, How todo
The FollowMeButton can be created in Interface Builder by adding a UIButton and changing it's class to FollowMeButton or in code using the custom initialiser:
[self.view addSubview:[[FollowMeButton alloc] initWithTwitterAccount:#"chrismaddern" atOrigin:CGPointMake(205, 248) isSmallButton:YES]];
Two size modes are available controlled by setting isSmallButton in the initialiser or by later change the isSmall property of the object.

Resources