Make subform visible by making a selection in a dropdown

[donn44157260]

I am trying to make a subform visible if the user select yes from a dropdown.

I have the following setup on the form

Subfom1 -Dropdown with 3 entries, Select, No Yes.

Subform2 – Text field for users to enter test

The code I have been trying to use is

“execEvent("change"); In the Initialize event and in the change event I was using

if (xfa.event.change == "5") {

    subform2.presence = "visible";

}

When I try the form, nothing happens,. There is no errors in the Javascript debugger

Any ideas would be helpful.

I am running Adobe LiveCycle Designer 8.05.2073.1.374024

Any ideas would be helpful

Chomp

[Magus069]

If your default value is set to select you can write in the change event

if (this.rawValue == 5)

[donn44157260]

Thanks Robert;

I did as you suggested and I still cannot get the field to become visible. I tried a couple of things.

(1) In the Initialize Event - execEvent("change");

In the Change Event - if (this.rawValue == "1") {

why_subform.presence = "visible";

}

The subform did not become visible nor was there an error in the javascript debugger

(2) Took execEvent("change"); out of the Initialize Event.

In the Change Event - if (this.rawValue == "1") {

why_subform.presence = "visible";

}

The subform did not become visible nor was there an error in the javascript debugger

Initialize Event blank

Added to the Change Event

if (this.rawValue == "1") {

    why_subform.presence = "hidden";

}

if (this.rawValue == "2") {

    why_subform.presence = "visible";

}

The subform did not become visible nor was there an error in the javascript debugger

(3) Added execEvent("change"); to the Initialize Event

The subform did not become visible nor was there an error in the javascript debugger

Preference defaults are set to – Preview = Interactive Form

                                                 XDP Preview Format – Adobe 8 (Dynamic) XML Form

[Magus069]

if you have specified values in your dropdownlist,

i think you have to check if the value is 1 as an integer not a string