想开启android.provider.Setrings下的action以打开相关设置页面,在iyu中点击事件中如下编写,虽未报错,但不能跳转。请问是否为参数的错误,该如何实现呢?
javanew(intent,"android.content.Intent","String","android.provider.Settings.ACTION_ACCESSIBILITY_SETTINGS")
java(null,activity,"android.app.Activity.startActivity","android.content.Intent",intent)
恳请指教,多谢!
javanew(intent,"android.content.Intent","String","android.provider.Settings.ACTION_ACCESSIBILITY_SETTINGS")
java(null,activity,"android.app.Activity.startActivity","android.content.Intent",intent)
恳请指教,多谢!