Reputation: 411
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
Upvotes: 0
Views: 1465
Reputation: 17503
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()});
}
});
Upvotes: 2
Reputation: 780
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.
Upvotes: 0
Reputation: 11835
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.
Upvotes: 0