CreateCricle:输入构件,及与之做碰撞检查的构件类别,进行碰撞检查。 public static IList<ElementId> FindIntersectsElementIdsByBuiltInCategory(Document doc, Element element, BuiltInCategory builtI…
反过来,我们也可以使用“按ID选择”,输入ID在Revit中选中相应的物体。如果要选中多个物体,同样可以用逗号分隔,一次输入多个ID。 我们知道,Revit中除了普通的工作视图,其它视图里也可以对图元进行操作,典型的如明细表视图。其实,在明细表视图中选中一行,等于是选中了对应的图元。 在Dynamo中,我们有Element.Id运算器,可...
1 review OS: Language: Description This tool will extract all the element IDs from all Autodesk® Revit® elements and store them in a shared parameter called "Element_ID" under each element. This will help in coordination and other purposes. ...
ToElementIds() - 获取匹配过滤器的第一个图元或图元ID。Obtain the first Element or ElementId tha...
Element elem = doc.get_Element(id); if(elemisWall) { if(reference.Proximity < (beamLen + tolerate))//Proximity接近,即与射线原点的距离。 { sel.Elements.Add(elem); } } } trans.Commit(); returnResult.Succeeded; } } url:http://greatverve.cnblogs.com/p/FindReferencesWithContextByDirection....
在Revit中使用“移动”工具时,移动对象的图元ID在移动后会发生更改。 原因:“取消关联”选项处于选中状态。 解决方案: 要避免在移动/移动图元时图元ID发生更改,请取消选中“移动”功能中的“取消连接”选项: 注意:选中“取消连接”后,“移动”工具将删除图元并创建新实例来删除约束。这将更改图元ID。...
.Select(q => q.Id)); } uidoc.Selection.SetElementIds(sameSc); } For Joel who said “I’d love to have a button to turn off annotations for all links in a view!”, the Revit API is getting closer to being able to do this but isn’t quite all the way there yet. ...
Where to find out details about the revit.exe file data transfer between components No 29-Aug-22 When editing the boundaries of the materials created with the split face option, it gives an error, and if an element is deleted, all the materials disappear in Revit. Edit boundaries ...
It will be a great improvement if we can also make a section box by selecting an object in linked RVT. 回覆 Jon Smith (發佈者) | 十月04, 2021 Hello Chau - this functionality is already in the app. To select a linked element, start the command with no object selected, pick the ...
From the documentation, we can see that GetParentId results InvalidElementId for root elements, so we do a first run through to find all root elements. Then, for each get again and recursively get the childs again. It’s not easy to show a hierarchical result, so let’s create aTreeVie...