Question

是否可以通过编程方式为StateListDrawable更改“状态”？

<selector xmlns:android="http://schemas.android.com/apk/res/android" android:exitFadeDuration="@android:integer/config_shortAnimTime">

<item
    android:state_checked="true"
    android:drawable="@drawable/picker_circle_selected"/>

<item
    android:state_checked="false"
    android:drawable="@drawable/picker_circle_today" />

StateListDrawable backgroundDrawable = (StateListDrawable) ContextCompat.getDrawable(getContext(), R.drawable.picker_selector);

我在selectDrawable(int index)上尝试了.....“ addState()”和“ StateListDrawable”。但没有任何效果。

默认显示“ state_checked = false”可绘制对象。当用户点击此可绘制对象时，它的状态更改为“ state_checked = true”。可以通过编程方式更改其状态吗？

Answer 1

您可以使用LayerDrawable

更改统计信息的颜色

DrawableContainerState drawableContainerState = (DrawableContainerState) backgroundDrawable.getConstantState();
Drawable[] children = drawableContainerState.getChildren();
LayerDrawable selectedItem = (LayerDrawable) children[0];
LayerDrawable unselectedItem = (LayerDrawable) children[1];


selectedItem.setColor(Color.Black); // changing to black color

Answer 2

尝试一下

// call your view
View view = LayoutInflater.from(viewGroup.getContext())
            .inflate(yourlayout, viewGroup, false);

    ColorDrawable colorDrawableSelected = new ColorDrawable(context.getResources().getColor(R.color.white));
    StateListDrawable stateListDrawable = new StateListDrawable();
    stateListDrawable.addState(new int[]{android.R.attr.state_selected}, colorDrawableSelected);
    stateListDrawable.addState(StateSet.WILD_CARD, null);// set the StateListDrawable as background of the view
    if (android.os.Build.VERSION.SDK_INT < android.os.Build.VERSION_CODES.JELLY_BEAN) {
        view.setBackgroundDrawable(stateListDrawable);
    } else {
        view.setBackground(stateListDrawable); 


 then call like this: view.setSelected etc

如何以编程方式更改StateListDrawable？

2 个答案: