问题描述
假设我有一个扩展 ViewGroup 的类
Suppose i have a class which extends ViewGroup
public class MapView extends ViewGroup
它包含在布局中 map_controls.xml 像这样
It is included in the layout map_controls.xml like this
<com.xxx.map.MapView android:id="@+id/map" android:background="@drawable/address" android:layout_width="fill_parent" android:layout_height="fill_parent"> </com.xxx.map.MapView>
如何从 AttributeSet 检索构造函数中的属性?假设背景字段中的可绘制对象.
How do i retrieve properties in the constructor from AttributeSet ? Let's say the drawable in the background field.
public MapView(Context context, AttributeSet attrs) { }
推荐答案
一般情况下,你这样做:
In the general case, you do like this:
public MapView(Context context, AttributeSet attrs) { // ... int[] attrsArray = new int[] { android.R.attr.id, // 0 android.R.attr.background, // 1 android.R.attr.layout_width, // 2 android.R.attr.layout_height // 3 }; TypedArray ta = context.obtainStyledAttributes(attrs, attrsArray); int id = ta.getResourceId(0 /* index of attribute in attrsArray */, View.NO_ID); Drawable background = ta.getDrawable(1); int layout_width = ta. getLayoutDimension(2, ViewGroup.LayoutParams.MATCH_PARENT); int layout_height = ta. getLayoutDimension(3, ViewGroup.LayoutParams.MATCH_PARENT); ta.recycle(); }
注意 attrsArray 中元素的索引如何影响.但是,在您的特定情况下,使用 getter 一样好,就像您自己发现的那样:
Pay attention to how the indexes of the elements in in attrsArray matter. However, in your particular case, it works just as good to use the getters, like you discovered yourself:
public MapView(Context context, AttributeSet attrs) { super(context, attrs); // After this, use normal getters int id = this.getId(); Drawable background = this.getBackground(); ViewGroup.LayoutParams layoutParams = this.getLayoutParams(); }
这是因为您在 com.xxx.map.MapView 上拥有的属性是 View 基类在其构造函数中解析的基本属性.如果您想定义自己的自己的属性,请查看这个问题和出色的答案:使用 XML 声明一个自定义的 android UI 元素
This works because the attribute you have on com.xxx.map.MapView are basic attributes that the View base class parses in its constructor. If you want to define your own attributes, take a look at this question and the excellent answer: Declaring a custom android UI element using XML