RIR doesn't create any floor if one fails?

Hi guys,

I was creating revit floors and Add Floor node fails when one of the boundaries fails? Any idea why I am getting this error. It doesn’t create any floor.

All bundled in one transaction. It may highlight the bad boundary if not can you cycle through and find the offending Boundary? Hard to tell what the error is without seeing the curve.

How can I cycle through it?

I ended up doing with a Python script btw.

You can create a list of indices to pick in the value picker or if you have human loaded you can use the handy Item Selector component and set it to cycle or sequence.

I have about 500 curveloops to loop through. I don’t think I’ll be able to make it. :frowning:

In that case looping through the branches, or sub list of the branches to narrow it down. Take half, see if it fails, take the other half, repeat until you find the right branch.

I was talking about this with Kike the other day, he’s going to add some info to help identify the bad actors.

1 Like

Yes, I think it would be great to see the index of the branch/item that fails the transaction.

ok, lots of duplicate rooms and ones that show a closed curve but are like the one below (a separation line)

Sending the details in a private message.

image

That would be great to see this feature implemented. Im working with creating elements through RiR alot and this used to happen often when managing a lot of curves and shapes, its not easy sometimes to figure out which one of unput data caused thetransaction failure.

As we talked with @Japhy , connecting a surface in the boundary input seems to work too. And it doesn’t abort the transaction.

2 Likes

It works as expected with surfaces.

1 Like

Thank you for the solution with surfaces, it worked in my case.

I’ve isolated a case where the boundary doesn’t work with curves, if you want to debug it.
Same curves work OK with python. The error seems to be caused by that vertex.
floor curves error.gh (9.9 KB)

1 Like

Thanks for reporting. It doesn’t like the radiused curve segments at those locations.

1 Like