GUILayout.Button does not respond to clicking

Question

I decided to make a beautiful component. I decided to make a button that adds a new field. A piece of code:

using System.Collections; using System.Collections.Generic; using UnityEngine; using UnityEditor; [CustomEditor(typeof(Starter))] public class StarterEditor : Editor { /// <summary> /// Компонент /// </summary> private Starter starter; /// <summary> /// Получение компонента при активации /// </summary> public void OnEnable() { starter = (Starter)target; } /// <summary> /// Изменение вида компонента /// </summary> public override void OnInspectorGUI() { int count = starter.managers.Count; EditorGUILayout.IntField("Размер", count); //Если уже есть Менеджеры в памяти, то вывести их список if (starter.managers.Count > 0) { foreach( var manage in starter.managers) { int index = starter.managers.IndexOf(manage); starter.managers[index] = (Manager.ManagerBase)EditorGUILayout.ObjectField("Менеджер", manage, typeof(Manager.ManagerBase), false); } } else { /* Если елементов в списке нету */ EditorGUILayout.LabelField("Нет елементов в списке"); } //Кнопка добавления нового менеджера if (GUILayout.Button("Добавить менеджер")) { Manager.ManagerBase newManager = (Manager.ManagerBase)EditorGUILayout.ObjectField("Менеджер", null, typeof(Manager.ManagerBase), false); } } }

But whatever I enter into the if () {} construction is not executed. All connected, and inherited a class from Editor. I will be glad to any help.

Inherited from EditorWindow and EditorWindow to OnGUI ?

M. Green M. Green 3,987 one 6 15 · Accepted Answer · 2018-09-02T20:25:20

The OnInspectorGUI method runs every frame while the inspector window is active. Accordingly, if (GUILayout.Button("Добавить менеджер")) also checked every frame, and the drawing of the field that is inside the condition occurs only in the frame in which the condition returns true.

As a result, when the button is pressed, the newManager variable is created, into which the result of the field drawing is written. After that, you do nothing with the variable and it is simply deleted when you exit the method, as a result, the button is not pressed in the next frame, you didn’t do anything with the result and the field is not drawn again.

If you had inserted a banal Debug.Log("Я тутъ"); inside the button click Debug.Log("Я тутъ"); , then they themselves would have perfectly seen in the console that pressing the button is being worked out, like everything that is written inside the condition.

The output is, for example, a list of your Manager.ManagerBase , in which, by clicking GUILayout.Button("Добавить менеджер") you will add a new instance of the class. And in the main body of the OnInspectorGUI method OnInspectorGUI you can draw a field through the foreach around the field for each instance that is in this sheet.

Now you draw directly from target, then you can directly add to it, so that you don’t have to rewrite anything.

From your next question, I realized that Manager.ManagerBase is inherited from ScriptableObject , which means that you will not be able to create it using new() and you will have to use something like this to save the new instance to the file system:

 if (GUILayout.Button("Добавить менеджер")) { Manager.ManagerBaseobj = ScriptableObject.CreateInstance<Manager.ManagerBase>(); AssetDatabase.CreateAsset(obj, "путь до папки"); //здесь вы можете добавить только что созданный ассет в список на отрисовку }

GUILayout.Button does not respond to clicking

1 answer 1

More articles: