我试图按如下所示设置按钮的可见性：
public Bundle setActivityState(Bundle bundle){
startBtn = (Button) findViewById(R.id.startSensorsBtn);
startBtn.setVisibility(
getVisibilityState(bundle, PersistanceConstants.START_BTN_STATE)
);
return bundle;
}
public int getVisibilityState(Bundle bundle, String keyName){
if (bundle.getInt(keyName) == View.VISIBLE){
return View.VISIBLE;
} else if (bundle.getInt(keyName) == View.INVISIBLE){
return View.INVISIBLE;
} else if (bundle.getInt(keyName) == View.GONE){
return View.GONE;
}
return 0;
}
但是我得到了错误：
Must be one of: View.VISIBLE, View.INVISIBLE, View.GONE less... (Ctrl+F1)
Reports two types of problems:
- Supplying the wrong type of resource identifier. For example, when calling Resources.getString(int id), you should be passing R.string.something, not R.drawable.something.
- Passing the wrong constant to a method which expects one of a specific set of constants. For example, when calling View#setLayoutDirection, the parameter must be android.view.View.LAYOUT_DIRECTION_LTR or android.view.View.LAYOUT_DIRECTION_RTL.
在通话时
getVisibilityState(bundle, PersistanceConstants.START_BTN_STATE)
我不知道该如何解决。我知道这是期望给定的一组值，但我所知道的只是将其传递int给它。在这里可以做什么？
当您知道自己在做什么时，可以通过以下方式在本地取消对Android Studio的检查：
//noinspection ResourceType
例如，
//noinspection ResourceType
startBtn.setVisibility(bundle.getInt(PersistanceConstants.START_BTN_STATE));
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我来说两句