前言
平时我们对View的显示判断都是用简要的方式去判断，那么，究竟是用view.isShown()去判断还是用view.
getVisibility() == View.VISIBLE 判断好呢？其实可以来看看源码
源码
- isShow()
/**
* Returns the visibility of this view and all of its ancestors
*
* @return True if this view and all of its ancestors are {@link #VISIBLE}
*/
public boolean isShown() {
View current = this;
//noinspection ConstantConditions
do {
if ((current.mViewFlags & VISIBILITY_MASK) != VISIBLE) {
return false;
}
ViewParent parent = current.mParent;
if (parent == null) {
return false; // We are not attached to the view root
}
if (!(parent instanceof View)) {
return true;
}
current = (View) parent;
} while (current != null);
return false;
}
可以看出注释
只有当view本身以及它的所有祖先们都是visible时，isShown()才返回TRUE。
- View.VISIBLE
/**
* This view is visible.
* Use with {@link #setVisibility} and <a href="#attr_android:visibility">{@code
* android:visibility}.
*/
public static final int VISIBLE = 0x00000000;
而平常我们调用if(view.getVisibility() == View.VISIBLE)只是对view本身而不对祖先的可见性进行判断。
总结
其实精炼总结成一句话就是，要判断用户能否看见某个view，应使用isShown()。
