Leahnaya/FModel
1
0
Fork 0
mirror of https://github.com/4sval/FModel.git synced 2026-08-26 03:14:39 -05:00
Code Issues Actions Packages Projects Releases Wiki Activity

Fix output pose layer detection to use AnimGraphNode_Root Name="AnimGraph"

Browse Source 
The output pose layer is now correctly identified by finding the node
whose Name property is "AnimGraph" (stored on AnimGraphNode_Root),
instead of incorrectly matching any node with "Root" in ExportType.

Co-authored-by: LoogLong <86428208+LoogLong@users.noreply.github.com>
This commit is contained in:
copilot-swe-agent[bot]
2026-03-02 06:31:15 +00:00
parent 1ba3afb483
commit c1962c77be
2 changed files with 5 additions and 5 deletions
Download Patch File Download Diff File Expand all files Collapse all files

4
FModel/ViewModels/AnimGraphViewModel.cs
View File

@@ -215,9 +215,9 @@ public class AnimGraphViewModel
/// </summary> /// </summary>
private static string GetLayerName(List<AnimGraphNode> nodes, int index) private static string GetLayerName(List<AnimGraphNode> nodes, int index)
{ {
// Look for a prominent node type to name the layer // The final output pose layer contains an AnimGraphNode_Root with Name="AnimGraph"
var rootNode = nodes.FirstOrDefault(n => var rootNode = nodes.FirstOrDefault(n =>
n.ExportType.Contains("Root", StringComparison.OrdinalIgnoreCase)); n.Name.Equals("AnimGraph", StringComparison.OrdinalIgnoreCase));
if (rootNode != null) if (rootNode != null)
return "AnimGraph"; return "AnimGraph";

6
FModel/Views/AnimGraphViewer.xaml.cs
View File

@@ -67,10 +67,10 @@ public partial class AnimGraphViewer
if (_viewModel.Layers.Count == 0) if (_viewModel.Layers.Count == 0)
return; return;
// Show only the final output pose layer (AnimGraph) initially // Show only the final output pose layer (AnimGraph) initially.
// The root node of the output layer is an AnimGraphNode_Root with Name="AnimGraph".
var outputLayer = _viewModel.Layers.FirstOrDefault(l => var outputLayer = _viewModel.Layers.FirstOrDefault(l =>
l.Nodes.Any(n => n.ExportType.Contains("Root", StringComparison.OrdinalIgnoreCase) || l.Nodes.Any(n => n.Name.Equals("AnimGraph", StringComparison.OrdinalIgnoreCase)))
n.ExportType.Contains("Result", StringComparison.OrdinalIgnoreCase)))
?? _viewModel.Layers[0]; ?? _viewModel.Layers[0];
var layersToShow = new[] { outputLayer }; var layersToShow = new[] { outputLayer };