on 11-24-2006 7:59 AM
Hi,
I have created a Submit button and a table UI.
Requirement is
1) Intial deployment button and table should be visible (its working fine)
2) if i click on Submit Button, Table UI should not be visible (its working fine)
3) if once again i click on Submit Button if Table should be visible ( How to implement this).
i.e on Action of submit button if table is already visible it should invisible and viceversa.
Please help.
Regards,
Bharath
Hi,
1. Create a context attribute say visibility of type WDVisibility in ur view context.
2. Bind this attribute to ur table ui element "visibility" property.
3. In doInit() method of view set visibility attribute to visible like this:
wdContext().currentContextElement().setVisibility(WDVisibility.VISIBLE);
On Action Submit:
if(wdContext().currentContextElement().getVisibility() == WDVisibility.VISIBLE){
wdContext().currentContextElement().setVisibility(WDVisibility.NONE);
}
else{
wdContext().currentContextElement().setVisibility(WDVisibility.VISIBLE);
}
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hi,
Thanks for response,I have awarded the points.
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Hi Bharath,
Action button code
if(tableFlag = true)
{
tableFlag = false;
}
else
tableFlag = true;
DoModifyView Code:
if(tableFlag = true)
{
wdContext.currentContextelement.setVisible (WDVisibility.VISIBLE);
tableFlag = true;
}
else
{
wdContext.currentContextelement.setVisible (WDVisibility.NONE);
tableFlag = false;
}
Declare it static in bottom
tableFlag = false;
I try to do work around like this
Regards, Suresh KB
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I guess you binded a Visibility attribute of context to the UI visible properties.
com.sap.ide.webdynpro.uielementdefinitions.Visibility
if (wdContext.currentContextElement().getVisAtrr() == WDVisibility.VISIBLE)
wdContext.currentContextElement().setVisAtrr(WDVisibility.NONE)
else
wdContext.currentContextElement().setVisAtrr(WDVisibility.VISIBLE)
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