I Think you could store temporary the continuation-ID and call the
continuation by pushing the back-button?!
Maybe!
Dirk
Hello,
I've in an application something like a cycle of an order which is divided into 3
forms. Now I have to implement a back button which shows the previous page with his
state.
Does someone know an easy and elegant way to implement this?
I've read something with createPageLocal(), but does this also work with sendview()?
Examples of my source you find below.
Thanks
Mike
//source in flowscript
function order( args_p ) {
jxform( "orderForm",
"order",
null,
null,
"request");
}
function orderForm( form_p )
{
var model = form_p.getModel();
do {
form_p.sendView( "orderPage1.form", validateOrderPage1Form );
}
while( !model.displayNextPage );
do {
form_p.sendView( "orderPage2.form", validateOrderPage2Form );
}
while( !model.displayNextPage );
do {
form_p.sendView( "orderPage3.form", validateOrderPage3Form );
}
while( !model.displayNextPage );
doTheOrder();
if ( model.errorCode == 0 ) {
form_p.sendView( "oK.form", validateOKForm );
}
else {
form_p.sendView( "noK.form", validateNOKForm );
}
form_p.finish();
}
function validateOrderPage1Form( form_p ){ //validateOrderPage2Form &
//validateOrderPage3Form are equal
var model = form_p.getModel();
var action = form_p.getSubmitId();
if ( action == "next" ) {
if (doSomeValidation()) {
model.displayNextPage = true;
}
else {
model.displayNextPage = false;
}
}
<!-- source of the form -->
<xf:submit id="next" continuation="forward">
<xf:label><i18n:text i18n:catalogue="buttons">NEXT</i18n:text></xf:label>
</xf:submit>
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