在新的 Unity 输入系统中如何映射到 KeyCode.JoystickButton0？

Question

我正在为 Android TV 和 Fire TV 开发一个 Unity 项目，该项目使用 v2019.3 中发布的新输入系统。

Back                    KeyCode.Escape
Select (D-Pad Center)   KeyCode.JoystickButton0
Left (D-Pad)            KeyCode.LeftArrow
Right (D-Pad)           KeyCode.RightArrow
Up (D-Pad)              KeyCode.UpArrow
Down (D-Pad)            KeyCode.DownArrow

除了 Select/D-Pad 中心之外，我已经在新系统中使用以下绑定路径成功映射了所有内容：

Escape [Keyboard]
Right Arrow [Keyboard]
Left Arrow [Keyboard]
Up Arrow [Keyboard]
Down Arrow [Keyboard]

如果我映射到 Any Key [Keyboard] 并实现以下代码作为回调，它显示为 key: JoystickButton0，这与亚马逊的文档相匹配，但在新系统中不作为选项存在在键盘绑定路径下：

public void DebugKey(InputAction.CallbackContext context)
{
    foreach(KeyCode vKey in System.Enum.GetValues(typeof(KeyCode))){
         if(Input.GetKey(vKey)){
             Debug.Log ("key: " + vKey);
         }
    }
}

我尝试了 Gamepad、Android Gamepad、Joystick 和 Android Joystick 下的各种按钮，但均未成功。另一个奇怪的事情是 Center/D-Pad 在没有任何绑定的情况下在 UI 按钮上工作正常。

使用新输入系统映射到旧 KeyCode.JoystickButtonX 命名约定的正确方法是什么？

谢谢！

Answer 1

在新系统中，设备被读取为HID，KeyCode没有用，所以如果你想读取摇杆按钮的状态，你可以这样做：

using UnityEngine.InputSystem;


public Joystick joy;

void Start()
{
     joy = Joystick.current;
}

void FixedUpdate()
{
    var button2 = joy.allControls[2].IsPressed();
    if (button2)
    {
        Debug.Log("Button2 of current joystick is pushed");
    }
}

如果您想映射按钮 0（在旧输入系统中），现在是按钮 1 或触发器

var button1 = joy.allControls[1].IsPressed();

或

var button1 = joy.trigger.IsPressed();

而且你可以测试更多棒的按钮..

void Start()
{
    var ListOfJoys = Joystick.all;
     joy = Joystick.current;//here current joy is ListOfJoys[1]
     otherjoy = ListOfJoys[0];
}


    var Buttons = joy.allControls;


    if ((Buttons[2] as ButtonControl).wasPressedThisFrame)
    {
        Debug.Log("b2 pressed during this frame");
    }
    if ((Buttons[2] as ButtonControl).wasReleasedThisFrame)
    {
        Debug.Log("b2 released during this frame");
    }
    if (joy.trigger.wasPressedThisFrame)
    {
        Debug.Log("trig or button1 pressed during this frame");
    }
    if (joy.trigger.wasReleasedThisFrame)
    {
        Debug.Log("trig or button1 released during this fram");
    }
    if (otherjoy.trigger.wasPressedThisFrame)
    {
        Debug.Log("otherjoy trigger");
    }

另一种方法是使用新系统的映射，我已经将动作press Only Test1映射到stick

的button3

简化代码，您可以混合使用事件和直接测试：

//Newinputsystem is a name of class generated 
private Newinputsystem playerAction;

void Awake()
{
    playerAction = new Newinputsystem();
    playerAction.Player.Test1.performed += ctx => FireTest();
}

void FixedUpdate()
{
    if (playerAction.Player.Test1.triggered)
    {
        Debug.Log("fire!!!");
    }
}

public void FireTest()
{
    Debug.Log("i am in firetest");
}

private void OnEnable()
{
    playerAction.Enable();
}
private void OnDisable()
{
    playerAction.Disable();
}

因此您可以添加新的动作 test2，它将被按钮 3 的动作 release only 触发...

要显示 HID 设备，请转到菜单 Window/analysis/input 调试器

你得看看你的设备（我的是hotas）

然后您单击它并在选项卡项 HID 描述符中单击

所以我想你已经在项目设置中选择了你的设备：

Answer 2

为了将使用 Unity 的 Amazon Fire TV 控制器与新的输入系统完全映射，您需要一个自定义设备。

使用此脚本，您可以使所有按钮正常工作。也可以调高音量、调低音量和静音。但我不认为覆盖此按钮是个好主意。

using System.Linq;
using UnityEngine;
using UnityEngine.InputSystem;
using UnityEngine.InputSystem.Controls;
using UnityEngine.InputSystem.Layouts;
using UnityEngine.InputSystem.LowLevel;
using UnityEngine.InputSystem.Utilities;
using UnityEngine.Scripting;
 
#if UNITY_EDITOR
using UnityEditor;
#endif
 
[Preserve]
public struct AftvRemoteDeviceState : IInputStateTypeInfo
{
    public FourCC format => new FourCC('A', 'G', 'C');
 
    //Mapped Here
    [InputControl(name = "dPadUpButton", layout = "Button", bit = 19, displayName = "Dpad Up")]
    [InputControl(name = "dPadRightButton", layout = "Button", bit = 22, displayName = "Dpad Right")]
    [InputControl(name = "dPadDownButton", layout = "Button", bit = 20, displayName = "Dpad Down")]
    [InputControl(name = "dPadLeftButton", layout = "Button", bit = 21, displayName = "Dpad Left")]
    //[InputControl(name = "backButton", layout = "Button", bit = 4, displayName = "Back Button")]
    [InputControl(name = "menuButton", layout = "Button", bit = 82, displayName = "Menu Button")]
    //Non Maped
    [InputControl(name = "selectButton", layout = "Button", bit = 23, displayName = "Select Button")]
    [InputControl(name = "rewindButton", layout = "Button", bit = 89, displayName = "Rewind Button")]
    [InputControl(name = "playPauseButton", layout = "Button", bit = 85, displayName = "Play Pause Button")]
    [InputControl(name = "fastForwardButton", layout = "Button", bit = 90, displayName = "Fast Forward Button")]
    public uint buttons;
}
 
#if UNITY_EDITOR
[InitializeOnLoad] // Call static class constructor in editor.
#endif
[Preserve]
[InputControlLayout(displayName = "Aftv Remote", stateType = typeof(AftvRemoteDeviceState))]
public class AftvRemoteDevice : InputDevice
{
#if UNITY_EDITOR
    static AftvRemoteDevice()
    {
        Initialize();
    }
#endif
 
    [RuntimeInitializeOnLoadMethod]
    private static void Initialize()
    {
        InputSystem.RegisterLayout<AftvRemoteDevice>
            (
                matches: new InputDeviceMatcher()
                .WithInterface("Android")
                .WithDeviceClass("AndroidGameController")
                .WithProduct("Amazon Fire TV Remote")
            );
    }
 
    public ButtonControl dPadUpButton { get; private set; }
    public ButtonControl dPadRightButton { get; private set; }
    public ButtonControl dPadDownButton { get; private set; }
    public ButtonControl dPadLeftButton { get; private set; }
    public ButtonControl selectButton { get; private set; }
    public ButtonControl rewindButton { get; private set; }
    public ButtonControl playPauseButton { get; private set; }
    public ButtonControl fastForwardButton { get; private set; }
    //public ButtonControl backButton { get; private set; }
    public ButtonControl menuButton { get; private set; }
    public bool SelectButtonDown { get; set; }
    public bool RewindButtonDown { get; set; }
    public bool PlayPauseButtonDown { get; set; }
    public bool FastForwardButtonDown { get; set; }
 
    protected override void FinishSetup()
    {
        base.FinishSetup();
        dPadUpButton = GetChildControl<ButtonControl>("dPadUpButton");
        dPadRightButton = GetChildControl<ButtonControl>("dPadRightButton");
        dPadDownButton = GetChildControl<ButtonControl>("dPadDownButton");
        dPadLeftButton = GetChildControl<ButtonControl>("dPadLeftButton");
 
        rewindButton = GetChildControl<ButtonControl>("rewindButton");
        playPauseButton = GetChildControl<ButtonControl>("playPauseButton");
        fastForwardButton = GetChildControl<ButtonControl>("fastForwardButton");
 
        //backButton = GetChildControl<ButtonControl>("backButton");
        menuButton = GetChildControl<ButtonControl>("menuButton");
        selectButton = GetChildControl<ButtonControl>("selectButton");    
    }
 
    public static AftvRemoteDevice current { get; private set; }
 
    public override void MakeCurrent()
    {
        base.MakeCurrent();
        current = this;
    }
 
    protected override void OnRemoved()
    {
        base.OnRemoved();
        if (current == this)
            current = null;
    }
 
    #region Editor
#if UNITY_EDITOR
    [MenuItem("Tools/AftvRemote/Create Device")]
    private static void CreateDevice()
    {
        InputSystem.AddDevice<AftvRemoteDevice>();
    }
    [MenuItem("Tools/AftvRemote/Remove Device")]
    private static void RemoveDevice()
    {
        var customDevice = InputSystem.devices.FirstOrDefault(x => x is AftvRemoteDevice);
        if (customDevice != null)
            InputSystem.RemoveDevice(customDevice);
    }
#endif
    #endregion
}

位值与 来自 AndroidAPI 的常数值相同。您可以查看列表 Here 您只需要将此脚本包含在项目中，设置您的输入操作和输入操作映射。

由于某种原因，播放器输入组件有问题。但是对于 UI 导航来说，这很好用，它也适用于输入调试器。

在新的 Unity 输入系统中如何映射到 KeyCode.JoystickButton0？

How do you map to KeyCode.JoystickButton0 in the new Unity Input System?

unity3d

android-tv

amazon-fire-tv