android Listview中设置enable状态，显示点击的item

1、listview 点击事件

    lvLeftMemu.setOnItemClickListener(new OnItemClickListener() {

@Override
public void onItemClick(AdapterView<?> arg0, View arg1, int position,
long arg3) {
mCurrentPos=position;
myMenuAdapter.notifyDataSetChanged();

}

2、Adapter中判断

   //区分list中item的点击状态
if (position==mCurrentPos) {
tvMenu.setEnabled(true);
}else {
tvMenu.setEnabled(false);
}

3、xml定义select选择器

  <?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android" >
    <item android:state_enabled="true" android:color="#fff"/>
    <item  android:color="#f00"/>
</selector>