为什么xposed无法找到方法?

时间:2019-05-18 06:24:28

标签: android xposed xposed-framework

  XposedHelpers.findAndHookMethod("com.android.server.input.InputManagerService",
            lpparam.classLoader, "nativeInjectInputEvent", int.class, lpparam.classLoader.loadClass("android.view.InputEvent"), int.class, int.class, int.class, int.class, int.class, new XC_MethodHook() {
                @Override
                protected void beforeHookedMethod(MethodHookParam param) throws Throwable {


                    XposedBridge.log("lzf called nativeInjectInputEvent:" + param.args[3]);
                    XposedBridge.log("uid is :" + TargetUid + " " + param.args[3]);
                    if ((Integer) param.args[3] == TargetUid) {
                        XposedBridge.log("here:" + param.args[3]);
                        param.args[3] = 0;
                    }
                }
            });

TargetUid 是我程序的UID,我希望他通过系统的权限检测。但我收到此错误:

  

java.lang.NoSuchMethodError:com.android.server.input.InputManagerService#nativeInjectInputEvent(int,android.view.InputEvent,int,int,int,int,int)#exact

这是源代码的“ InputManagerService”。link

我在做什么错了?

0 个答案:

没有答案