Exposing a function to the inspector

Question

I wish to expose a script's function to the inspector, but I have no clue how to. I found [this][1] here, which was a start, but it isn't clear how to finish.

My code for the class:

 [System.Serializable]
     class MoveAction : UnityEvent<MonoBehaviour>, IUIAnimationEvent
     {
         #region Move
         public MoveAction()
         {
 
         }
         //to ensure only one mover coroutine can be active.
         IEnumerator moveRoutine = null;
         #region Solution 2: using fields and not parameters
         Transform from;
         Transform to;
         float overTime;
         public delegate void UIchain(MonoBehaviour mono);
         public event UIchain NEXT_FUNCTION;
         
         public MoveAction(Transform from, Transform to, float overTime, IUIAnimationEvent chain)
         {
             this.from = from;
             this.to = to;
             this.overTime = overTime;
 
         }
         public void Move(MonoBehaviour mono)
         {
             if (moveRoutine != null)
             {
                 mono.StopCoroutine(moveRoutine);
             }
             moveRoutine = _Move(from, to, overTime, mono);
             mono.StartCoroutine(moveRoutine);
             Invoke(mono);
         }
         IEnumerator _Move(Transform from, Transform to, float overTime, MonoBehaviour mono)
         {
             Vector2 original = from.position;
             float timer = 0.0f;
             while (timer < overTime)
             {
                 float step = Vector2.Distance(original, to.position) * (Time.deltaTime / overTime);
                 from.position = Vector2.MoveTowards(from.position, to.position, step);
                 timer += Time.deltaTime;
                 yield return null;
             }
             if(NEXT_FUNCTION != null)
             {
                 NEXT_FUNCTION(mono);
             }
         }

This is a lot of code, but the interesting part is the Move(MonoBehaviour mono) function. I want to expose that to the editor. How? [1]: http://answers.unity3d.com/questions/998183/select-monobehaviour-function-from-inspector.html

Answer 1

Answer by hexagonius · Feb 07, 2017 at 10:51 PM

you cannot expose a function (it's called method in C#). you can however expose a unity event (like in your posted question). all you need is

 [System.Serializable]
      public class MoveAction : UnityEvent<MonoBehaviour>{}

and in a different class

 public MoveAction moveAction;

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agiro · Feb 08, 2017 at 10:34 AM 0

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@hexagonius Thanks, it worked - as long as my function doesn't accept more parameters than 1. However, I have a collection of functions where I use at least 3 parameters. How to make it work for 3 or more parameters?

Answer 2

Answer by UnityCoach · Feb 08, 2017 at 10:41 AM

If what you want is a button that you can click in the inspector to do some editor work, you need to make a custom inspector.

 [CustomEditor(typeof(YourClassName))]
 [CanEditMultipleObjects] // only if you handle it properly
 public class YourClassNameEditor : UnityEditor.Editor
 {
     public override void OnInspectorGUI()
     {
         if (GUILayout.Button("DO THAT", EditorStyles.miniButton))
         {
             (YourClassName)this.target).YourMethod (parameters);
         }
         DrawDefaultInspector ();
     }
 }

You will also need to handle Undo then.

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Answer 3

Answer by ChristianLaLarsson · Oct 13, 2019 at 07:20 PM

I found the answer I was looking for here: https://gamedev.stackexchange.com/questions/136996/unity-exposing-a-function-to-the-inspector?newreg=f40cf113944e4884b47fb09e2f16dc41

what you are looking for is the Unity Event class. See this example below. This will appear in the editor exactly like your screenshot.

 using UnityEngine;
 using UnityEngine.Events;
 using System.Collections;
 
 public class InvokeOnAwake : MonoBehaviour {
 
     public UnityEvent invokeMethod;//set in editor
 
     void Awake(){
         invokeMethod.Invoke();
     }
 
 }

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Exposing a function to the inspector

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