我在我的游戏中捕捉截图时,球员死亡。我有以下代码来捕捉屏幕截图。
RenderTexture rt = new RenderTexture (800, 600, 24);
MainCamera.targetTexture = rt;
Texture2D texture = new Texture2D (800, 600, TextureFormat.RGB24, false);
MainCamera.Render ();
RenderTexture.active = rt;
texture.ReadPixels (new Rect (0, 0, 800, 600), 0, 0);
MainCamera.targetTexture = null;
RenderTexture.active = null;
Destroy (rt);
byte[] bytes = texture.EncodeToPNG ();
Directory.CreateDirectory (Application.persistentDataPath + "/GameOverScreenShot");
File.WriteAllBytes (Application.persistentDataPath + "/GameOverScreenShot" + "/DiedScreenShot.png", bytes);
我正在使用以下代码获取保存的屏幕截图。
byte[] bytes = File.ReadAllBytes (Application.persistentDataPath +"/GameOverScreenShot" + "/BirdDiedScreenShot.png");
Texture2D texture = new Texture2D (800, 600, TextureFormat.RGB24, false);
RectOffset tempOffset = new RectOffset (5, 5, 5, 5);
texture.filterMode = FilterMode.Trilinear;
texture.LoadImage (bytes);
Sprite sprite = Sprite.Create (texture, new Rect (0, 0, 800, 400), new Vector2 (0.5f, 0.0f), 2.0f);
ScreenShot_Image.GetComponent<Image> ().sprite = sprite;
现在,我想分享这个截图的android应用程序。根据我的研究,我有以下代码,但它是返回空白图像。
//instantiate the class Intent
AndroidJavaClass intentClass = new AndroidJavaClass("android.content.Intent");
//instantiate the object Intent
AndroidJavaObject intentObject = new AndroidJavaObject("android.content.Intent");
//call setAction setting ACTION_SEND as parameter
intentObject.Call<AndroidJavaObject>("setAction", intentClass.GetStatic<string>("ACTION_SEND"));
//instantiate the class Uri
AndroidJavaClass uriClass = new AndroidJavaClass("android.net.Uri");
//instantiate the object Uri with the parse of the url's file
string destination = Application.persistentDataPath + "/GameOverScreenShot" + "/DiedScreenShot.png";
AndroidJavaObject uriObject = uriClass.CallStatic<AndroidJavaObject>("parse","file://"+destination);
//call putExtra with the uri object of the file
intentObject.Call<AndroidJavaObject>("putExtra", intentClass.GetStatic<string>("EXTRA_STREAM"), uriObject);
//set the type of file
intentObject.Call<AndroidJavaObject>("setType", "image/*");
//instantiate the class UnityPlayer
AndroidJavaClass unity = new AndroidJavaClass("com.unity3d.player.UnityPlayer");
//instantiate the object currentActivity
AndroidJavaObject currentActivity = unity.GetStatic<AndroidJavaObject>("currentActivity");
//call the activity with our Intent
currentActivity.Call("startActivity", intentObject);
我应该在这方面改变什么??请帮助,前进谢谢
2条答案
按热度按时间fruv7luv1#
只需调用
takeScreenShotAndShare()
来拍摄屏幕快照并共享它。如果您已经有了要共享的图像,只需调用StartCoroutine(shareScreenshot(path));
并传入图像的路径/位置。这只支持png图像。要共享jpeg,请更改至
整个代码:
643ylb082#
在Android 8+中,此方法无效。我们需要使用FileProvider。
Google更改了使用FileProvider的共享方式,使其更加安全,我们不能将文件对象直接共享给任何其他应用程序。我们需要使用FileProvider并授予读取URI权限,使其可供任何其他应用程序访问。
我使用了
androidx.core.content.FileProvider
,因为所有库都迁移到AndroidX。您也可以使用android.support.v4.content.FileProvider
。完整的Post by Suneet Agrawal is here。我只提供了关键步骤。