Question

如何添加一个自定义的按钮状态

Answer 1

<?xml version="1.0" encoding="utf-8"?>
<resources>
<declare-styleable name="food">
<attr name="state_fried" format="boolean" />
<attr name="state_baked" format="boolean" />
</declare-styleable>
</resources>

public FoodButton(Context context, AttributeSet attrs) {
super(context, attrs);
}

private static final int[] STATE_FRIED = {R.attr.state_fried};
private static final int[] STATE_BAKED = {R.attr.state_baked};

private boolean mIsFried = false;
private boolean mIsBaked = false;

public void setFried(boolean isFried) {mIsFried = isFried;}
public void setBaked(boolean isBaked) {mIsBaked = isBaked;}
Then override function "onCreateDrawableState":

@Override
protected int[] onCreateDrawableState(int extraSpace) {
final int[] drawableState = super.onCreateDrawableState(extraSpace + 2);
if (mIsFried) {
mergeDrawableStates(drawableState, STATE_FRIED);
}
if (mIsBaked) {
mergeDrawableStates(drawableState, STATE_BAKED);
}
return drawableState;
}

文件 "res/drawable/food_button.xml":
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res/com.mydomain.mypackage">
<item
app:state_baked="true"
app:state_fried="false"
android:drawable="@drawable/item_baked" />
<item
app:state_baked="false"
app:state_fried="true"
android:drawable="@drawable/item_fried" />
<item
app:state_baked="true"
app:state_fried="true"
android:drawable="@drawable/item_overcooked" />
<item
app:state_baked="false"
app:state_fried="false"
android:drawable="@drawable/item_raw" />
</selector>