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Java 具有2种颜色的ListView不交错_Java_Android_Listview - Fatal编程技术网

Java 具有2种颜色的ListView不交错

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

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

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      }