How to open and close wiQuery Dialogs? - java

I'm trying out wiQuery to see if it suits my needs, but I've run into problems with very basic stuff. Consider the following, where I try to control when a Dialog opens and closes, using its open() and close() methods:
HTML:
<input type="submit" wicket:id="open" value="Open dialog"/>
<div wicket:id="dialog">
<input type="submit" wicket:id="close" value="Close"/>
</div>
Java:
final Dialog dialog = new Dialog("dialog");
add(new Link("open") {
#Override
public void onClick() {
dialog.open();
}
});
dialog.add(new Link("close") {
#Override
public void onClick() {
dialog.close();
}
});
add(dialog);
Thing is, the above doesn't work.
The only way I've got the dialog to actually open & close from my code is by calling setAutoOpen() with either true or false, but it seems strange is this is the only way. (That method's Javadoc says "Sets if this window opens autmatically after the page is loaded." so it clearly should be reserved for a different purpose.)
What's the right way of opening and closing wiQuery Dialogs dynamically in your code?

I have been using the last 2 weeks and I have a similar problem. Try using an AjaxLink this way:
AjaxLink openingLink = new AjaxLink("open")
{
#Override
public void onClick(AjaxRequestTarget target)
{
// Do something with model
target.addComponent(content);
dialog.open(target);
}
};

Related

Processing single input field with ajax in Apache Wicket

I have a form with several input fields and one special field, that I want to process with ajax. The thing is, that I want to process only that field after the AjaxLink has been clicked. Without processing of the whole form. I want to access the value of that input field in the method onSubmit of the AjaxLink. Is that possible? If yes, then how?
Regards,
Mateusz
By default AjaxLink does not submit data/forms. AjaxSubmitLink and AjaxButton do!
For your use case you can AjaxRequestAttributes and send "dynamic extra parameters". I'm on my mobile and I cannot give you an example at the moment but the idea is to construct a simple JSON object with a key being the request parameter name and value the forn element's value.
Google these keywords!
If you can't manage to do it then add a comment and I will update my answer as soon as I can!
Here is a sample code. Beware I've written it completely here, so it might have a typo or two!
add(new AjaxLink("customSubmitLink") {
#Override public void onClick(AjaxRequestTarget target) {
int aFieldValue = getRequest().getRequestParameters().getParameterValue("aField").toInt();
// do something with aFieldValue
}
#Override protected void updateAjaxAttributes(AjaxRequestAttributes attrs) {
super.updateAjaxAttributes(attrs);
attrs.getDynamicExtraParameters().add("return {\"aField\": jQuery('#aFormField').val()});
}
});
One way to solve this would be to put that 'special' field with its 'special' link to a second Form and then use CSS to visually position the 'special' field like it is inside the main Form.
Something like this:
Form<Void> mainForm = new Form<Void>("main-form") {
#Override
protected void onSubmit() {
super.onSubmit();
}
};
add(mainForm);
// ... populate the main form
Form<Void> secondForm = new Form<Void>("second-form");
add(secondForm);
final Model<String> specialModel = Model.of();
secondForm.add(new TextField<>("special-field", specialModel));
secondForm.add(new AjaxButton("special-button") {
#Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
// ... process the special field value
}
});
And in the markup, as usual:
<form wicket:id="main-form">
... main form content
</form>
<form wicket:id="second-form">
<label>Special field: <input class="special-field" wicket:id="special-field"></label>
<button wicket:id="special-button">Special button</button>
</form>
And then style that .special-field class with position: absolute; top: ... or something like that.
The solution is not very elegant, it's more of a hack. It will create some confusion for a person who would have to read this later. But it may work if the trick with CSS is possible.
It's actually even easier than what rpuch suggested.
Just nest your forms and make sure the AjaxLink only submits the second form:
<form wicket:id="form">
<div wicket:id="dateTimeField"></div>
<form wicket:id="secondForm">
<input wicket:id="text" />
<a wicket:id="secondSubmit">submit2</a>
</form>
<a wicket:id="submit">submit</a>
</form>
Form secondForm= new Form("secondForm");
form.add(secondForm);
final IModel<String> textModel = Model.of("");
TextField<String> text = new TextField<>("text", textModel);
secondForm.add(text);
AjaxSubmitLink secondSubmit = new AjaxSubmitLink("secondSubmit", secondForm) {
#Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
super.onSubmit(target, form);
logger.info("textMod: " + textModel.getObject());
}
};
secondForm.add(secondSubmit);
The second form will be rendered as a div but will have the functionality that you desire. However the second form will also be submitted when you submit the outer form.

Wicket AjaxLink in Modal Window doesn´t work

I am trying to add an AjaxLink inside a ModalWindow. This AjaxLink is used to do some stuff like deleting something off the database and finally close the ModalWindow.
I added the ModalWindow accordording to the Wicket examples: Link to examples. But this doesn´t work.
My MainPage:
public class EventPanel extends Panel {
// some stuff happens here, the constructor accepts the eventModel
final ModalWindow modal;
add(modal = new ModalWindow("modal"));
modal.setCookieName("modal-1");
modal.setPageCreator(new ModalWindow.PageCreator() {
public Page createPage() {
// Use this constructor to pass a reference of this page.
return new DeleteEventWindow(eventModel, modal);
}
});
modal.setCloseButtonCallback(new ModalWindow.CloseButtonCallback() {
public boolean onCloseButtonClicked(AjaxRequestTarget target) {
// Change the passValue variable when modal window is closed.
return true;
}
});
// Add the link that opens the modal window.
add(new AjaxLink<Void>("showModalLink") {
#Override
public void onClick(AjaxRequestTarget target) {
modal.show(target);
}
});
}
Modal Window:
public class DeleteEventWindow extends WebPage {
public DeleteEventWindow(final IModel<Event> model,
final ModalWindow window) {
// some stuff happens
// this link doesn´t work
add(new AjaxLink<Void>("closeOK") {
#Override
public void onClick(AjaxRequestTarget target) {
// Just a print to console for debugging
System.out.println("nooo");
window.close(target);
}
});
}
}
ModalWindow HTML
<html>
<head>
<title>Modal Content Page</title>
</head>
<body>
<!-- some other fields output --!>
<a wicket:id="closeOK">close</a><br/>
</body>
</html>
The ModalWindow itself works fine, also the link is rendered. But if I click onto it, the onClick function doesn´t seem to be triggered. I also tried a normal Link, this works fine..
I also found this question : stackoverflow question, but I am using JQuery 1.9.1..
Your code is not giving enough information; you have the modal window added to an item (row Item I assume), but the link to show the modal is added to the panel.

Wicket: Preventing form submission by a TextField within a Wizard step

Question relates to Wicket 1.6
I have a wizard step, which includes a Textfield component. When I press the Enter key, this is being handled by the default button of the Wizard bar ('Next'), and it advances to the next step in the Wizard. I don't want this to happen. When I hit Enter on the Textfield I just want the value to be updated, but remain on the same page.
I tried overriding the onBeforeRender() method of my Wizard class, which as you can see sets the default button of the containing form to null. However this now results in the 'Prev' button being triggered when I hit Enter, so the wizard goes back to the previous step.
public class ConfigurationWizard extends Wizard {
....
#Override
protected void onBeforeRender()
{
super.onBeforeRender();
Component buttonBar = getForm().get(BUTTONS_ID);
if (buttonBar instanceof IDefaultButtonProvider)
{
getForm().setDefaultButton(null);
}
}
}
So the basic question is, how do I disable the default button behaviour of the Wizard?
My approach (with a nice Wicket behavior)
Usage
TextField<String> myField = new TextField<String>("myField", myModel());
myField.add(new PreventSubmitOnEnterBehavior());
Behavior
public class PreventSubmitOnEnterBehavior extends Behavior
{
private static final long serialVersionUID = 1496517082650792177L;
public PreventSubmitOnEnterBehavior()
{
}
#Override
public void bind( Component component )
{
super.bind( component );
component.add( AttributeModifier.replace( "onkeydown", Model.of( "if(event.keyCode == 13) {event.preventDefault();}" ) ) );
}
}
This has nothing to do with the wizard buttons.
The TextField <input> is doing a form submit when the Enter key is pressed. This is standard behaviour for the <input> element.
The solution is to catch the Enter key press for the <input> and prevent the default behaviour
This bit of javascript magic does the trick for me:
<script type="text/javascript">
$(document).ready(function() {
$("#gridDiv").delegate("input","keypress",function(e){
if(e.originalEvent.keyCode == 13){
e.preventDefault();
}
});
});
</script>
where 'gridDiv' is the id of the <div> containing the TextField
I prefer another approach:
I use AjaxButtons for every button needed, with the specific submit code in the overrided onSubmit():
AjaxButton linkSubmit = new AjaxButton("linkSubmit")
#Override
public void onSubmit(AjaxRequestTarget target, Form form) {
super.onSubmit();
// Submit code goes here....
// ...
setResponsePage(new NewPage());
}
#Override
public void onError(AjaxRequestTarget target, Form form) {
}
};
My form doesn't need a "onSubmit()" method.
And the markup doesn't have any submit buttons. All buttons are coded like this:
With this approach you don't need to mess with javascript codes. The page simply will do nothing if you press Enter. You'll have to click your buttons to submit each one.
Hope this can help you.

How to disable / change style of wicket button link in onClick()

In a Wicket app, I have a bunch of <button> elements to which I'm attacking a Link component. Now in the onClick() method of the component I want to disable or change the style of the button. How can I do that? Calling setEnabled(false) has no effect.
Repeated uses of onClick() are operating on the same object in memory. If you're not using Ajax, you can still maintain some state in an anonymous subclass of Link. Then, you can use onBeforeRender() and onComponentTag() to change how it is displayed each time.
Link<Void> link = new Link<Void>("myLink") {
private String customCSS = null;
private boolean customEnabled = true;
public void onClick() {
if (/* test to determine disabled */) {
customCSS = "disabled";
customEnabled = false;
} else {
customCSS = null;
customEnabled = true;
}
}
#Override
protected void onComponentTag(ComponentTag tag) {
super.onComponentTag(tag);
if (customCSS != null)
tag.put("class", customCSS);
}
#Override
public boolean isEnabled() {
return super.isEnabled() && customEnabled;
}
};
AttributeModifiers (or other behaviors) aren't good for this case because, if you add them in the onClick() method, they will begin stacking on the same link for each click - since they are maintained as part of the Link's state.
Your Link can keep track of all manner of state, allowing your onClick() method to enable/disable/change/etc with repeated clicks.
You can also override onBeforeRender(), isVisible(), and other methods that are run each time the link is displayed on the page. The constructor, onConfigure(), and others are run just once, regardless of how many times you click the button.
I don't think this is an entirely good idea in Wicket. Of course it could be done by trickery, but it's far simpler to either:
Override the isEnabled() method to return a value derived from the model of the form/component.
Attach an AttributeModifier when you create the component, and use a model for it which returns a value derived as above.
Whichever you choose, the principle is to let Wicket "pull" rendering information in rather than pushing it explicitly.
The answer provided by Michael Borgwardt is nearly correct.
The problem is that you use Link. Disabled Links use <span> instead of
<a>/<button> and are surrounded with <em> by default. Using Button
component will set 'disabled' attribute in the element.
I would like to add, that you need to use HTML button element instead of <a> (link). Original answer can be counfusing, because Link and Button also exist in Wicket.
I think AjaxCallDecorator should be the class you need to use to disable/change style of the button.
The problem is that you use Link. Disabled Links use <span> instead of <a>/<button> and are surrounded with <em> by default.
Using Button component will set 'disabled' attribute in the element.
Take a look at SimpleAttributeModifier and AttributeAppender. Depending on your actual requirements one of those should do the trick. SimpleAttributeModifier adds or replaces an attribute of any HTML-Tag that has a prepresentation in wicket (replaces the css class), while AttributeAppender appends to the attributes (adds another css class). This should work for enabling/disabling buttons as well but I haven't tried that.
Example:
Label label = new Label("id", "Some silly text.")
add(label);
label.add(new SimpleAttributeModifier("class", "my-css-class");
For Ajax you'll have to add the component to the target as well.
More detailed example:
Java code:
import org.apache.wicket.behavior.AttributeAppender;
import org.apache.wicket.behavior.SimpleAttributeModifier;
import org.apache.wicket.markup.html.WebMarkupContainer;
import org.apache.wicket.markup.html.WebPage;
import org.apache.wicket.markup.html.form.Button;
import org.apache.wicket.markup.html.form.Form;
import org.apache.wicket.markup.html.link.Link;
import org.apache.wicket.model.Model;
public class DemoPage extends WebPage {
public DemoPage() {
Form form = new Form("form");
add(form);
final WebMarkupContainer wmc = new WebMarkupContainer("greenText");
form.add(wmc);
form.add(new Link("redLink"){
#Override
public void onClick() {
wmc.add(new SimpleAttributeModifier("class", "redText"));
}});
final Button boldButton = new Button("boldButton"){
#Override
public void onSubmit() {
wmc.add(new AttributeAppender("class", true, new Model<String>("boldText"), " "));
}};
form.add(boldButton);
Link disabler = new Link("buttonDisabler") {
#Override
public void onClick() {
boldButton.add(new AttributeAppender("disabled", true, new Model<String>("disabled"), " "));
}
};
form.add(disabler);
}
}
corresponding HTML:
<html>
<head>
<style>
.redText {
color: red;
}
.greenText {
color: green;
}
.boldText {
font-weight: bold;
}
</style>
</head>
<body>
<form wicket:id="form">
<div class="greenText" wicket:id="greenText">This is Green.</div><br />
Make it red<br />
<input type="submit" wicket:id="boldButton" value="Make it bold" /><br />
Disable the button
</form>
</body>
</html>

GWT open page in a new tab

I am developing GWT application and I use
com.google.gwt.user.client.Window.open(pageUrl, "_blank", "");
to open new page. And it opens in a new tab when called, for example, directly after button click.
But I decided to do some validations on server before opening new page and placed the call to the mentioned above method to the
public void onSuccess(Object response) {
}
And it starts to open pages in new window instead of new tab (this is true only for Chrome, other browsers still open it in a new tab).
Can anybody help me?
I built a small example to illustrate the issue:
button.addClickHandler(new ClickHandler() {
public void onClick(ClickEvent event) {
Window.open("http://www.google.com/", "_blank", "");
MySampleApplicationServiceAsync serviceAsync = GWT.create(MySampleApplicationService.class);
serviceAsync.getMessage("Hello, Server!", new AsyncCallback() {
public void onFailure(Throwable caught) {
Window.alert("ERROR");
}
public void onSuccess(Object result) {
Window.open("http://www.bing.com/", "_blank", "");
}
}
);
}
});
Firefox(3.6.8) opens both pages in new tabs.
Chrome(6.0) opens "google.com" in new tab and "bing.com" in new window
Opera(10.10) opens in new tabs.
IE(8.0) opens both in new Windows.
I marked igorbel 's answer as the only correct cos I haven't found any proper way to specify the same behaviour in all situations.
I used this code and it works for me in google chrome and mozilla firefox 3.6.8 browsers
If you want to open a page in new window you should write code as
Window.open("www.google.com","_blank","enabled");
If you want to open a page in new tab you should write code as
Window.open("www.google.com","_blank","");
I am not sure you are going to be able to control this the way you want. The problem is that browsers can decide when to open windows and when to open tabs. For example, firefox has the option: "Open new windows in new tabs instead". And don't forget the browsers that don't support tabs (yes, those do still exist).
Since this is such a problematic aspect of the user experience, my recommendation would be to reconsider your design. Is it really that important for you application to differentiate between opening a new tab and opening a new window?
This code works for me:
Before calling the Async method keep a reference to a new window with empty parameters.
At onSuccess() method set the URL of the window.
Button someButton = new Button("test");
SelectionListener<ButtonEvent> listener = new SelectionListener<ButtonEvent>()
{
public void componentSelected(ButtonEvent ce)
{
final JavaScriptObject window = newWindow("", "", "");
someService.doSomething(new AsyncCallback()
{
public void onSuccess(Object o)
{
setWindowTarget(window, "http://www.google.com/");
}
});
}
}
someButton.addSelectionListener(listener);
private static native JavaScriptObject newWindow(String url, String name, String features)/*-{
var window = $wnd.open(url, name, features);
return window;
}-*/;
private static native void setWindowTarget(JavaScriptObject window, String target)/*-{
window.location = target;
}-*/;
Found at:
http://groups.google.com/group/google-web-toolkit/browse_thread/thread/574b3b828271ba17
Interesting thing,
chrome will open page in new tab in case if you put window.open(...) instruction into the body of the click handler implementation.
For example:
Button someButton = new Button("test",
new ClickHandler() {
public void onClick(ClickEvent event) {
Window.open(...);
}
});
And a page will be opened in the separate window in case if I will include any Async. request into the mentioned code:
Button someButton = new Button("test",
new ClickHandler() {
public void onClick(ClickEvent event) {
someService.doSomething(new AsyncCallback() {
void onSuccess(Object o) {
Window.open(...);
}
...
});
}
});
The way Chrome looks at it, calling Window.open() is like trying to open a pop-up window in the user's face. That's frowned upon and will trigger the built-in pop-up blocker. Following a link, according to Chrome, should be the result of a user clicking on a good old anchor tag with an href attribute. But here lies the answer you're looking for: you can show a link to the user and change the link target on the fly. That would qualify as a 'proper' link in Chrome's world.
This code works for me:
public static native String getURL(String url)/*-{
return $wnd.open(url,
'target=_blank')
}-*/;

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