Java 具有2种颜色的ListView不交错
我正在尝试用两种颜色做一个列表视图,白色和灰色交错。在getView方法上的适配器中,我执行以下代码:Java 具有2种颜色的ListView不交错,java,android,listview,Java,Android,Listview,我正在尝试用两种颜色做一个列表视图,白色和灰色交错。在getView方法上的适配器中,我执行以下代码: if(position % 2 == 0){ v.setBackgroundColor(Color.WHITE); }else{ v.setBackgroundResource(R.color.light_grey_listas); } return v; 但在我的屏幕上,有时一组线的颜色是相同的。例如,3行为灰色背景色,或者当我在列表视图中导航时,该行会因错误的颜色而更改
if(position % 2 == 0){
v.setBackgroundColor(Color.WHITE);
}else{
v.setBackgroundResource(R.color.light_grey_listas);
}
return v;
但在我的屏幕上,有时一组线的颜色是相同的。例如,3行为灰色背景色,或者当我在列表视图中导航时,该行会因错误的颜色而更改
以下是要执行的步骤
步骤1.1)对奇数和偶数位置列表项使用两个选择器
艺术家列表背景颜色.xml
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item
android:state_selected="false"
android:state_pressed="false"
android:drawable="@color/grey" />
<item android:state_pressed="true"
android:drawable="@color/itemselected" />
<item android:state_selected="true"
android:state_pressed="false"
android:drawable="@color/itemselected" />
</selector>
步骤1.2)
艺术家\u列表\u背景\u alternate.xml
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item
android:state_selected="false"
android:state_pressed="false"
android:drawable="@color/sign_out_color" />
<item android:state_pressed="true"
android:drawable="@color/login_hover" />
<item android:state_selected="true"
android:state_pressed="false"
android:drawable="@color/login_hover" />
</selector>
步骤2)
colors.xml
<?xml version="1.0" encoding="utf-8"?>
<resources>
<color name="survey_toplist_item">#EFEDEC</color>
<color name="survey_alternate_color">#EBE7E6</color>
<color name="grey">#ffffff</color>
<color name="itemselected">#EDEDED</color>
<color name="login_hover">#E5F5FA</color>
<color name="sign_out_color">#e84040</color>
</resources>
@Override
public View getView(int position, View convertView, ViewGroup parent) {
View view = convertView;
if (view == null) {
view = lInflater.inflate(R.layout.listitem, parent, false);
}
if (position % 2 == 0) {
view.setBackgroundResource(R.drawable.artists_list_backgroundcolor);
} else {
view.setBackgroundResource(R.drawable.artists_list_background_alternate);
}
((TextView) view.findViewById(R.id.heading)).setText(data.get(position));
return view;
}
}
#EFEDEC
#EBE7E6
#ffffff
#埃德德
#E5F5FA
#e84040
@凌驾
公共视图getView(int位置、视图转换视图、视图组父视图){
视图=转换视图;
如果(视图==null){
视图=lInflater.flate(R.layout.listitem,父项,false);
}
如果(位置%2==0){
视图.setBackgroundResource(R.drawable.Artisters\u list\u backgroundcolor);
}否则{
视图.背景资源(R.drawable.Artisters\u list\u background\u alternate);
}
((TextView)view.findviewbyd(R.id.heading)).setText(data.get(position));
返回视图;
}
}
您可以在下面的链接中获得完整的描述
尝试以下getView()方法:
发布整个适配器代码,可能您使用的是holder模式,而没有正确处理它。
@Override
public View getView(int position, View convertView, ViewGroup parent) {
View view = convertView;
ViewHolder holder;
if (view == null) {
holder=new ViewHolder();
view = lInflater.inflate(R.layout.listitem, parent, false);
view.setTag(hoder);
}else{
holder = (ViewHolder) view.getTag();
}
if (position % 2 == 0) {
view.setBackgroundResource(R.drawable.artists_list_backgroundcolor);
} else {
view.setBackgroundResource(R.drawable.artists_list_background_alternate);
}
((TextView) view.findViewById(R.id.heading)).setText(data.get(position));
return view;
}
class ViewHolder {
//Declare here your listview variables }