Fix visibleIf attribute

Author: Antoine Lelievre

Assets/Examples/DefaultNodes/Nodes/VisibleIfAttributeNode.cs

@@ -0,0 +1,49 @@
+using System.Collections;
+using System.Collections.Generic;
+using UnityEngine;
+using GraphProcessor;
+using System.Linq;
+
+[System.Serializable, NodeMenuItem("Custom/VisibleIfAttributeNode")]
+public class VisibleIfAttributeNode : BaseNode
+{
+	public enum Test1
+	{
+		A,
+		B,
+		C,
+		D
+	}
+
+	public enum Test2
+	{
+		T1,
+		T2,
+		T3,
+	}
+
+	public Test1 t1;
+
+	[VisibleIf(nameof(t1), Test1.A)]
+	public float f1;
+	[VisibleIf(nameof(t1), Test1.B)]
+	public int f2;
+
+	[VisibleIf(nameof(t1), Test1.C)]
+	public string s1;
+	[VisibleIf(nameof(t1), Test1.C)]
+	public Test2 t2;
+
+	[Input(name = "In")]
+    public float                input;
+
+	[Output(name = "Out")]
+	public float				output;
+
+	public override string		name => "VisibleIfAttributeNode";
+
+	protected override void Process()
+	{
+	    output = input * 42;
+	}
+}

Assets/Examples/DefaultNodes/Nodes/VisibleIfAttributeNode.cs.meta

@@ -747,6 +747,8 @@ private void AddEmptyField(FieldInfo field, bool fromInspector)
 
 		void UpdateFieldVisibility(string fieldName, object newValue)
 		{
+			if (newValue == null)
+				return;
 			if (visibleConditions.TryGetValue(fieldName, out var list))
 			{
 				foreach (var elem in list)
@@ -839,6 +841,10 @@ protected VisualElement AddControlField(FieldInfo field, string label = null, bo
 			var element = new PropertyField(FindSerializedProperty(field.Name), showInputDrawer ? "" : label);
 			element.Bind(owner.serializedGraph);
 
+			element.RegisterValueChangeCallback(e => {
+				UpdateFieldVisibility(field.Name, field.GetValue(nodeTarget));
+			});
+
 			// Disallow picking scene objects when the graph is not linked to a scene
 			if (element != null && !owner.graph.IsLinkedToScene())
 			{