Re: FeedbackPanel rendering
Hi Andre; Thanks for your answer.
I didn't override form's onSubmit method.
From: Andrea Del Bene
To: users@wicket.apache.org
Sent: Thursday, August 25, 2011 11:42 AM
Subject: Re: FeedbackPanel rendering
Hi Lurtz,
what does your form do in its onSubmit method? Have you overriden it?
> Hi all;
>
>
> I have two panels.
>
>
> Panel one have the standart form with one Label + TextField (required field)
> + AjaxButton.
>
> Panel two have one FeedbackPanel.
>
>
> Ok, here is the the problem :
>
>
> I can reach Panel Two' FeedbackPanel from Panel One.
>
>
> So in AjaxButton
>
>
> AjaxButton submit = new AjaxButton("submit") {
> @Override
> protected void onError(AjaxRequestTarget target, Form
>form) {
> FeedbackPanel feedbackPanel = getFeedbackPanel(); //
>gets FeedBackPanel from Panel Two
> target.addComponent(feedbackPanel); // it renders, no
>problem
> target.addComponent(form);
> }
>
> @Override
> protected void onSubmit(AjaxRequestTarget target, Form
>form) {
> FeedbackPanel feedbackPanel = getFeedbackPanel(); //
>gets FeedBackPanel from Panel Two
> info("Saved"); // just an info message
> target.addComponent(form);
> target.addComponent(feedbackPanel); // no way
>
> }
> };
> When Validation error occurs (onError) Panel's two FeedbackPanel renders and
> shows the error. Great !
>
>
> But when there is no error (onSubmit)Panel's two FeedbackPanel doesn't render
> and it gives
>
>
> component-targetted feedback message was left unrendered. This could be
> because you are missing a FeedbackPanel on the page. Message:
> [FeedbackMessage message = "Saved", reporter = go, level = INFO]
>
>
> Any suggestions ?
>
> Regards.
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FeedbackPanel rendering
Hi all;
I have two panels.
Panel one have the standart form with one Label + TextField (required field) +
AjaxButton.
Panel two have one FeedbackPanel.
Ok, here is the the problem :
I can reach Panel Two' FeedbackPanel from Panel One.
So in AjaxButton
AjaxButton submit = new AjaxButton("submit") {
@Override
protected void onError(AjaxRequestTarget target, Form form) {
FeedbackPanel feedbackPanel = getFeedbackPanel(); // gets
FeedBackPanel from Panel Two
target.addComponent(feedbackPanel); // it renders, no
problem
target.addComponent(form);
}
@Override
protected void onSubmit(AjaxRequestTarget target, Form form)
{
FeedbackPanel feedbackPanel = getFeedbackPanel(); // gets
FeedBackPanel from Panel Two
info("Saved"); // just an info message
target.addComponent(form);
target.addComponent(feedbackPanel); // no way
}
};
When Validation error occurs (onError) Panel's two FeedbackPanel renders and
shows the error. Great !
But when there is no error (onSubmit)Panel's two FeedbackPanel doesn't render
and it gives
component-targetted feedback message was left unrendered. This could be because
you are missing a FeedbackPanel on the page. Message: [FeedbackMessage message
= "Saved", reporter = go, level = INFO]
Any suggestions ?
Regards.
Re: rendering component
setVisible(false) , actually renders component but disables it via style sheet
like
I don't want that.
I will try isVisibleInHierachy now.
Thanks.
From: Martin Grigorov
To: users@wicket.apache.org; Lurtz Nazgul
Sent: Thursday, August 11, 2011 1:32 PM
Subject: Re: rendering component
org.apache.wicket.Component.setVisibilityAllowed(boolean) is for that purpose.
if component.isVisibleInHierachy() returns false then the component is
not rendered.
What exactly is the problem with setVisible(false) ?
On Thu, Aug 11, 2011 at 12:59 PM, Lurtz Nazgul wrote:
> Hi,
>
> Is it possible not to render a component for security issues.
>
> I know i can use IAuthorizationStrategy for this but i want to to it manually
>
> I don't need just setVisible(false), i really need not to render that
> component.
>
> I don't know which method to override. Below code gives error.
>
> TextField nameField = new TextField("name", new
> PropertyModel(this,
> "name")) {
> @Override
> protected void onRender(MarkupStream markupStream) {
>
> }
> };
>
> Any way ?
>
> Thanks.
--
Martin Grigorov
jWeekend
Training, Consulting, Development
http://jWeekend.com
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rendering component
Hi,
Is it possible not to render a component for security issues.
I know i can use IAuthorizationStrategy for this but i want to to it manually
I don't need just setVisible(false), i really need not to render that
component.
I don't know which method to override. Below code gives error.
TextField nameField = new TextField("name", new
PropertyModel(this,
"name")) {
@Override
protected void onRender(MarkupStream markupStream) {
}
};
Any way ?
Thanks.
Re: flush ajaxRequestTarget.appendJavascript
I used AjaxSelfUpdatingTimerBehavior and it's done.
Thanks.
From: Lurtz Nazgul
To: users@wicket.apache.org
Sent: Wed, August 3, 2011 11:36:50 AM
Subject: flush ajaxRequestTarget.appendJavascript
Hi;
When i append jquery progressbar script to AjaxRequestTarget like below
for (int i = 0; i < 101; i++) {
ajaxRequestTarget.appendJavascript("$(document).ready(function()
{\n" +
"$(\"#progressbar\").progressbar({ value: " + i
+ " });\n" +
"});");
// other stuff
}
Wicket sends it to the server, but i need to flush that jquery script code in
order to work progressbar.
Any suggestions ?
Thanks.
flush ajaxRequestTarget.appendJavascript
Hi;
When i append jquery progressbar script to AjaxRequestTarget like below
for (int i = 0; i < 101; i++) {
ajaxRequestTarget.appendJavascript("$(document).ready(function()
{\n" +
"$(\"#progressbar\").progressbar({ value: " + i
+ " });\n" +
"});");
// other stuff
}
Wicket sends it to the server, but i need to flush that jquery script code in
order to work progressbar.
Any suggestions ?
Thanks.
Re: catch jquery event at AjaxEventBehavior
Hi ;
I used AbstractDefaultAjaxBehavior to catch jquery close event.
On jquery side i triggered event when jquery event window closed.
wicketAjaxGet('"+behavior.getCallbackUrl()+ "'");
It is little tricky.
Thanks.
________
From: Lurtz Nazgul
To: users@wicket.apache.org
Sent: Mon, August 1, 2011 9:57:30 AM
Subject: catch jquery event at AjaxEventBehavior
Hi;
I used jquery dialog. I can open and close jquery dialog box. I represent
jquery
dialog as a panel at wicket side.
My question is how can i catch jquery close event ?
I tested like below but it doesn't work
dialogConfirmation.add(new AjaxEventBehavior("close") {
@Override
protected void onEvent(AjaxRequestTarget target) {
System.out.println(" closed" );
}
}) ;
Thanks.
catch jquery event at AjaxEventBehavior
Hi;
I used jquery dialog. I can open and close jquery dialog box. I represent
jquery
dialog as a panel at wicket side.
My question is how can i catch jquery close event ?
I tested like below but it doesn't work
dialogConfirmation.add(new AjaxEventBehavior("close") {
@Override
protected void onEvent(AjaxRequestTarget target) {
System.out.println(" closed" );
}
}) ;
Thanks.
Re: dynamic DataTable
Any simple way ? Thanks. From: Wilhelmsen Tor Iver To: "users@wicket.apache.org" Sent: Thu, July 21, 2011 3:10:16 PM Subject: RE: dynamic DataTable > Since TextArea can contain any SELECT statement, Column names, count can >change > > every time. So I need a dynamic table that is compatible to view any SELECT > Statement result. Since you appear to have no need for performing updates or inserts, try ditching the DataTable in favour of using nested repeaters over the ResultSetMetaData column info and ResultSet rows instead. - Tor Iver - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
dynamic DataTable
Hi, How can we create a dynamic DataTable which has dynamic Columns? I want to create a page that has following components: - A TextArea for SQL Statement input (User will type any SELECT Sql here) - A Submit button to run the SQL - A Table to view the SQL Resultset. Since TextArea can contain any SELECT statement, Column names, count can change every time. So I need a dynamic table that is compatible to view any SELECT Statement result. How can I simply do this in Wicket? Thanks a lot, Regards,
Re: Forward to Bookmarkable
Hi Hans;
Thanks for your answer
But still url can be seen by the user.
I need to forward in order to hide the url from user.
Any idea ?
Thanks.
From: Hans Lesmeister
To: Wicket Users
Sent: Thu, July 21, 2011 12:35:21 PM
Subject: Re: Forward to Bookmarkable
You can setResponePage or throw one of the RestartResponse*-Exceptions
Regards
Hans
Am 21.07.11 11:19 schrieb "Lurtz Nazgul" unter :
> I also tried
>
> WebRequestCycle cycle = (WebRequestCycle) RequestCycle.get();
> ServletRequest httpRequest =
> cycle.getWebRequest().getHttpServletRequest();
> ServletResponse httpResponse =
> cycle.getWebResponse().getHttpServletResponse();
> ServletContext context = ((WebApplication)
> Application.get()).getServletContext();
> context.getRequestDispatcher("/mywelcome").forward(httpRequest,
> httpResponse);
>
> where mywelcome url is BookmarkablePage, in MyWebApplication.java
>
> this.mountBookmarkablePageWithUrlCoding("/mywelcome", Welcome.class);
>
> Thanks.
>
>
>
>
>
> From: Lurtz Nazgul
> To: users@wicket.apache.org
> Sent: Thu, July 21, 2011 12:15:19 PM
> Subject: Forward to Bookmarkable
>
> Hi;
>
> How can i forward user to a BookmarkablePage. I don't need redirection. I
> don't
>
> want user to see url.
>
>
> Thanks.
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Re: Forward to Bookmarkable
I also tried
WebRequestCycle cycle = (WebRequestCycle) RequestCycle.get();
ServletRequest httpRequest =
cycle.getWebRequest().getHttpServletRequest();
ServletResponse httpResponse =
cycle.getWebResponse().getHttpServletResponse();
ServletContext context = ((WebApplication)
Application.get()).getServletContext();
context.getRequestDispatcher("/mywelcome").forward(httpRequest,
httpResponse);
where mywelcome url is BookmarkablePage, in MyWebApplication.java
this.mountBookmarkablePageWithUrlCoding("/mywelcome", Welcome.class);
Thanks.
________
From: Lurtz Nazgul
To: users@wicket.apache.org
Sent: Thu, July 21, 2011 12:15:19 PM
Subject: Forward to Bookmarkable
Hi;
How can i forward user to a BookmarkablePage. I don't need redirection. I
don't
want user to see url.
Thanks.
Forward to Bookmarkable
Hi; How can i forward user to a BookmarkablePage. I don't need redirection. I don't want user to see url. Thanks.
Re: Generate ajax link
Thanks Martin; but it doesn't make sense to me, sorry.
Wicket generates below code for a simple counter increment example.
var wcall =
wicketAjaxGet('?wicket:interface=:2:panel1:okla::IBehaviorListener:0:',
function() {
}.bind(this), function() {
}.bind(this), function() {
return Wicket.$('okla') != null;
}.bind(this));
return !wcall;
Any suggestions ?
Thanks.
From: Martin Grigorov
To: users@wicket.apache.org
Sent: Thu, July 14, 2011 2:46:04 PM
Subject: Re: Generate ajax link
See how AjaxLink generates its markup.
On Thu, Jul 14, 2011 at 2:31 PM, Lurtz Nazgul wrote:
> I mean; generating ajax link format
>
> Instead of
>
>
>
> I want to generate ajax link on the fly
>
> Example :
> increment
>
> Any suggestions ?
>
> Thanks.
>
>
>
>
>
> From: Andrea Del Bene
> To: users@wicket.apache.org
> Sent: Thu, July 14, 2011 2:00:32 PM
> Subject: Re: Generate ajax link
>
> Hi,
>
> you can use Loop component to generate a dynamic number of links. For example:
>
>
>
> HTML code:
>
>
>
>
>
>
>
>
>
> Java code:
>
>
> List linkList = ...
> Loop loop = new Loop("itemList", linkList.size()) {
>@Override
>protected void populateItem(LoopItem item) {
>AbstractLink curLink = linkList.get(item.getIndex());
>item.add(curLink);
>}
>};
>
>
>
>> Hi all;
>>
>> I can generate normal links in to my dynamic menu
>>
>>
>> ...
>>
>> I wonder if i can generate those links in ajax link format ?
>>
>> Thanks.
>
>
> -
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--
Martin Grigorov
jWeekend
Training, Consulting, Development
http://jWeekend.com
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Re: Generate ajax link
I mean; generating ajax link format
Instead of
I want to generate ajax link on the fly
Example :
increment
Any suggestions ?
Thanks.
From: Andrea Del Bene
To: users@wicket.apache.org
Sent: Thu, July 14, 2011 2:00:32 PM
Subject: Re: Generate ajax link
Hi,
you can use Loop component to generate a dynamic number of links. For example:
HTML code:
Java code:
List linkList = ...
Loop loop = new Loop("itemList", linkList.size()) {
@Override
protected void populateItem(LoopItem item) {
AbstractLink curLink = linkList.get(item.getIndex());
item.add(curLink);
}
};
> Hi all;
>
> I can generate normal links in to my dynamic menu
>
>
> ...
>
> I wonder if i can generate those links in ajax link format ?
>
> Thanks.
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Generate ajax link
Hi all; I can generate normal links in to my dynamic menu ... I wonder if i can generate those links in ajax link format ? Thanks.
