所以我有自定义列表项,带有ListView按钮.按下时,按钮显示交替可绘制以向用户显示反馈.但是,当我点击该行时,每个按钮都显示按下状态,就像我点击它们一样.
如何让按钮显示其原始状态而不是state_pressed?
布局/列表项目:
<LinearLayout android:layout_width="0dp" android:layout_height="match_parent" android:layout_weight="1" android:orientation="horizontal" android:paddingBottom="10dp" android:paddingTop="10dp" android:descendantFocusability="blocksDescendants" > <LinearLayout android:layout_width="0dp" android:layout_height="match_parent" android:layout_weight="1" android:orientation="vertical" android:paddingLeft="10dp" android:paddingRight="10dp" android:gravity="center_vertical|left" > <TextView android:id="@+id/txtMain" android:layout_width="wrap_content" android:layout_height="wrap_content" android:singleLine="true" android:textAppearance="?android:attr/textAppearanceLarge" style="@style/PrimaryText" /> <TextView android:id="@+id/txtSub" android:layout_width="wrap_content" android:layout_height="wrap_content" android:singleLine="true" android:textAppearance="?android:attr/textAppearanceLarge" style="@style/SecondaryText" /> </LinearLayout> <ImageButton android:id="@+id/imbResponse" android:layout_width="wrap_content" android:layout_height="wrap_content" android:background="@null" android:focusable="false" android:duplicateParentState="false" android:src="@drawable/response_btn" android:contentDescription="@string/response" android:layout_marginLeft="10dp" android:layout_marginTop="5dp" android:layout_marginRight="10dp" android:layout_marginBottom="5dp" /> </LinearLayout>
绘制/ response_btn.xml:
<?xml version="1.0" encoding="utf-8"?> <selector xmlns:android="http://schemas.android.com/apk/res/android" > <item android:state_focused="true" android:drawable="@drawable/res_m" /> <item android:state_pressed="true" android:drawable="@drawable/res_m" /> <item android:state_focused="false" android:state_pressed="false" android:drawable="@drawable/res_alt_m" /> </selector>
我试图删除state_focused和state_pressed,state_focused.似乎该按钮从其父级获取state_pressed.
解决方法
我发现将android:clickable =“true”设置为父视图将阻止更改子视图状态.