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Modify the content of a container depending on a variable

Hi,

is it possible to change the content of a container depending on a variable's content?

I have a container containing several list boxes and if my variable is set to '1', I want to hide one of the listboxes.

Problem is, if I add a conditional show to the listbox, only the content of the listbox disappears, the listbox itself is still to be seen in the container. Please see small sample attached.

Thanks, Regards

Jens

1 Solution

Accepted Solutions
Digvijay_Singh

Hope you have added condition in layout tab/show in the list box of container. Its surprising if it is not working, I have QV12 version.

You can try unlinking the container list box by right clicking on listbox header, the outside box should not hide after that.

View solution in original post

11 Replies
Chanty4u
MVP
MVP

uncheck the objects in layout tab  size to data

Not applicable
Author

This is not working.

Not applicable
Author

Alternatively, I could use two different containers, put them in the same place and show/hide them depending on the value.

But i find this solution a bit ugly and would prefer something  better :-).

Digvijay_Singh

Add condition in the layout-show of the list box, it will completely vanished.

Not applicable
Author

No it won't. Only the content will disappear,

Digvijay_Singh

Check this sample, Change vHide input box value to 1 or 2.

Not applicable
Author

Only the standalone listbox disappears, but not the one that is part of container CT01.

If it works for you, might it be an issue with the QV version? I use V11 SR9.

Digvijay_Singh

Hope you have added condition in layout tab/show in the list box of container. Its surprising if it is not working, I have QV12 version.

You can try unlinking the container list box by right clicking on listbox header, the outside box should not hide after that.

Not applicable
Author

Then this is a new feature in V12. In V11 the behavior is like I described.

screen .jpg