I can only think of one way of doing this :
You should have two buttons, a 'submit button' one, and a 'confirm button'.
Everytime you hit the submit button, the form should be saved, to a
SessionState object
and return TO THE SAME page.
Then, on second page refresh ( after the hit button ), you can test if
the sessionstate object exists.
If it exists, you should show the confirm button, the one which will
actually store the data ( or send, or whatever )
.tml
<form>
<input type="submit" t:id="send" value="Save"/>
<t:if text="objectExists">
<input type="submit" t:id="confirm" value="confirm"/>
</t:if>
</form>
.java
@SessionState
private Object myObject
private Class successfullPage;
onSelectedFromSend() {
// populate myObject fields
successfullPage = null;
}
onSelectedFromConfirm() {
// actually saving data
successfullPage = WhatEverPage.class
}
Object onSuccess() {
return successfullPage;
}
Hello!
On a page I have a form which I want to redisplay after it has been
submitted. The user should be able to change/submit it many times until
he is satisfied with the result.
But when he clicks on the submit button, there is no visual response
that the submit happened. What I would expect is that the page blanks
and rebuilds after the submit is done.
Another idea is displaying a "Your query was submitted"-box in top of
the form, but after the second submit the effect would be the same: none.
Does anyone has a hint for me how I can visualize this?
Perhaps I could mark the changed values in a different color, but I do
not know how to access form fields which have been created with a loop
component.
TIA
Stephan
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