on 07-25-2008 6:14 AM
hello...
I m using interactive form. i placed a button in that form
i declared a method and also binded this method to the properties of interactive form. but how to give link between the button which i placed in form to the method which i created?
thanks in advance
renu
Put a standard button in adobe designer and choose SUBMIT radio button Control Type there.
Now come to ur view and go to Interactive Form properties.
Put ur required action under OnSubmit Event under properties
This action will be called on click of button in adobe form
Mandeep Virk
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i placed a button of type WebDynpro Active X at interactive form
when i click on this button, the perticular action is triggering
but the thing is here i want to display the values in table form
i m getting values and its displaying only one row. it should hav multiple rows. coz at backned i m having more than 10records
its showing only one record at table.
i couldnt resolve this now
how to display all records.....Here i check the option Repeat row for each data item" to display multiple records, even though its not displaying....
kindly put some light on this
Make a value node named Data Source of cardinality 1:1
Create another value node say Table Node inside Data Source of cardinality 0:N and singleton False (This will allow multiple child inside node).
This Data Source will be put under dataSource property of Interactive form UI Element. And Table Node will be actual data container for interactive form
Open the Adobe form layout. Drag "SubForm" UI Element from the library in to body pages of pdf layout. Now drag the "TableNode" node from "Data View" tab in to the subform.
Open the properties of Subform. Check the "Repeat Subform for each data item" check box
Refer this document for more information [Displaying Internal Table in Adobe Form|https://www.sdn.sap.com/irj/sdn/weblogs?blog=/pub/wlg/7762]
Inside your button refer and put this code
int size = wdContext.nodeTableNode().size() ;
IPrivate<view name>View.ITableElement elt ;
for(int i=0; i<size; i++){
elt = wdContext.createTableNode() ;
elt.setObj1();
elt.setObj2();
wdContext.nodeTableNode().addElement( elt );
}
Ask for further query
Mandeep Virk
Edited by: Mandeep Virk on Jul 25, 2008 1:25 PM
Hi renushree,
As you have said u have created a button, go to the properties of the button you have Events --> onAction --> where u need to create an Action (method), on clicking the button the Action will Trigger automatically. or you have give the method which have wrritten in Events onAction. This will resolve your Problem.
Regards,
Sharma.
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hi renu,
plz check out the following link which would help you solve your query
[use of interactive form|http://help.sap.com/saphelp_nw04s/helpdata/en/dc/f1783fe3263042e10000000a114084/frameset.htm]
[SAP Interactive Forms|https://www.sdn.sap.com/irj/sdn/adobe]
regards,
Murthy.
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First thing you have to create an action you will find that "Actions" tab when you open the iview. Now Select the button in layout mode and add the action to button's onAction property. Unless you don't assign an action to button will be disabled when you run your application.
P.S.
Actions are special purpose methods their life cycle contolled by framework. Code the action according to your requirements.
Jawed Ali
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