Question

当我在主页中时，我使用下面的代码显示突出显示的按钮，但是我的代码无效，所以我不明白我的xml代码中有什么问题。

请告诉我正确的方法吗？

<selector xmlns:android="http://schemas.android.com/apk/res/android">

    <item android:drawable="@drawable/button_pressed" android:state_activated="true"/>
 <!-- pressed -->
    <item android:drawable="@drawable/button_pressed" android:state_pressed="true"/>
 <!-- pressed -->
    <item android:drawable="@drawable/button_pressed" android:state_focused="true"/>
 <!-- focused -->
    <item android:drawable="@drawable/list_button" android:state_hovered="true"/>
 <!-- hovered -->
    <item android:drawable="@drawable/button_background"/>
 <!-- default -->

</selector>

Answer 1

试试这个，

<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">

<item android:drawable="@drawable/button_background" android:state_pressed="false"/>
<!-- default -->
<item android:drawable="@drawable/button_pressed" android:state_pressed="true"/>
<!-- pressed -->
</selector>

Answer 2

创建xml选择器文件

    <item android:drawable="@android:color/transparent" android:state_window_focused="false"/>
    <item android:drawable="@drawable/action_item_selected" android:state_pressed="true"/>
    <item android:drawable="@drawable/action_item_selected" android:state_focused="true"/>
    <item android:drawable="@android:color/transparent"/>

</selector>

Android按钮突出显示状态不起作用

2 个答案: