最近有一个列表效果,需要一个列表有多种布局,最终效果如下:
这个我也问了同事以及开发群里的朋友,居然都没得到最优的实现方式的回答,看来这种复杂列表的需求还是比较少的,我自己也走了一些弯路,把我几个实现的方式整理下,希望对于还不了解的朋友有所帮助。
实现方式1:(每次getView时重新inflate itemView,convertView没有复用,性能低,运行没问题)
private class MyAdapter extends BaseAdapter{ private List<Object> datas = Collections.EMPTY_LIST; public void setDatas(List<Object> datas) { if(datas == null){ datas = Collections.EMPTY_LIST; } this.datas = datas; notifyDataSetChanged(); } @Override public int getCount() { return datas.size(); } @Override public Object getItem(int position) { return datas.get(position); } @Override public long getItemId(int position) { return 0; } @Override public View getView(int position, View convertView, ViewGroup parent) { Object data = getItem(position); if(data instanceof Folder){ FolderViewHolder holder = null; if(convertView != null && convertView.getTag() instanceof FolderViewHolder){ //View与数据类型一致 holder = (FolderViewHolder) convertView.getTag(); }else{ convertView = mInflater.inflate(R.layout.listitem1, null); holder = new FolderViewHolder(convertView); convertView.setTag(holder); } holder.setData((Folder)data); }else{ FileViewHolder holder = null; if(convertView != null && convertView.getTag() instanceof FileViewHolder){ //View与数据类型一致 holder = (FileViewHolder) convertView.getTag(); }else{ convertView = mInflater.inflate(R.layout.listitem2, null); holder = new FileViewHolder(convertView); convertView.setTag(holder); } holder.setData((File)data); } return convertView; } } private class FolderViewHolder{ public TextView tvName; public FolderViewHolder(View itemView){ tvName = (TextView) itemView.findViewById(R.id.tvName); } public void setData(Folder data) { tvName.setText(data.name); } } private class FileViewHolder{ public TextView tvName; public FileViewHolder(View itemView){ tvName = (TextView) itemView.findViewById(R.id.tvName); } public void setData(File data) { tvName.setText(data.name); } }
实现方式2:(因为方式1不断inflate view,影响性能,于是考虑是否能尽可能重用已经inflate的view,于是添加了一个缓存,不过实际测试快速滑动或切换数据会显示异常,应该是AbsListView#RecycleBin缓存的原因,具体原因我后面理清了再添加,看别人的代码最痛苦了。。。)
private class MyAdapter extends BaseAdapter{ private List<View> folderViewCaches = new ArrayList<View>(5); private List<View> fileViewCaches = new ArrayList<View>(5); private List<Object> datas = Collections.EMPTY_LIST; public void setDatas(List<Object> datas) { if(datas == null){ datas = Collections.EMPTY_LIST; } this.datas = datas; notifyDataSetChanged(); } @Override public int getCount() { return datas.size(); } @Override public Object getItem(int position) { return datas.get(position); } @Override public long getItemId(int position) { return 0; } @Override public View getView(int position, View convertView, ViewGroup parent) { Object data = getItem(position); if(data instanceof Folder){ //文件夹,应该返回R.layout.listitem1对应的View FolderViewHolder holder = null; if(convertView != null && convertView.getTag() instanceof FolderViewHolder){ //View与数据类型一致 holder = (FolderViewHolder) convertView.getTag(); }else{ if(convertView != null){ //缓存到文件列表 fileViewCaches.add(convertView); convertView = null; } //从缓存里面取已从ListView移除的缓存(注释掉此部分代码显示正常) if(!folderViewCaches.isEmpty()){ for(View cache : folderViewCaches){ if(cache.getParent() == null){ //缓存的View已从listView里面移除 convertView = cache; holder = (FolderViewHolder) convertView.getTag(); folderViewCaches.remove(cache); break; } } } //还是没有,重新inflate if(convertView == null){ convertView = mInflater.inflate(R.layout.listitem1, null); holder = new FolderViewHolder(convertView); convertView.setTag(holder); } } holder.setData((Folder) data); }else{ //文件,应该返回R.layout.listitem2对应的View FileViewHolder holder = null; if(convertView != null && convertView.getTag() instanceof FileViewHolder){ //View与数据类型一致 holder = (FileViewHolder) convertView.getTag(); }else{ if(convertView != null){ //缓存到文件夹列表 folderViewCaches.add(convertView); convertView = null; } //从缓存里面取已从ListView移除的缓存(注释掉此部分代码显示正常) if(!fileViewCaches.isEmpty()){ for(View cache : fileViewCaches){ if(cache.getParent() == null){ //缓存的View已从listView里面移除 convertView = cache; holder = (FileViewHolder) convertView.getTag(); fileViewCaches.remove(cache); break; } } } //还是没有,重新inflate if(convertView == null){ convertView = mInflater.inflate(R.layout.listitem2, null); holder = new FileViewHolder(convertView); convertView.setTag(holder); } } holder.setData((File) data); } return convertView; } }
实现方式3:(最佳实现,运行正常)
后面仔细阅读ListView相关源码,才发现Adapter本身就支持不同的布局了,而且AbsListView#RecycleBin也支持不同类型的布局的缓存策略,RecycleBin.mViewTypeCount标示有多少种View类型。
我们需要做的就是重写Adapter的下面3个方法:
1.getViewTypeCount:
/** * 有多少种不同布局的View */ @Override public int getViewTypeCount() { return 2; }
2.getItemViewType
/** * 相应position对应的View类型 */ @Override public int getItemViewType(int position) { if(getItem(position) instanceof Folder){ return TYPE_FOLDER; }else{ return TYPE_FILE; } }
3.getView,通过判断对应position的类型,返回相应类型的view:
@Override public View getView(int position, View convertView, ViewGroup parent) { Object data = getItem(position); if(data instanceof Folder){ //TYPE_FOLDER,文件夹,应该返回R.layout.listitem1对应的View FolderViewHolder holder = null; if(convertView != null){ holder = (FolderViewHolder) convertView.getTag(); }else{ convertView = mInflater.inflate(R.layout.listitem1, null); holder = new FolderViewHolder(convertView); convertView.setTag(holder); } holder.setData((Folder) data); }else{ //TYPE_FILE,文件,应该返回R.layout.listitem2对应的View FileViewHolder holder = null; if(convertView != null){ holder = (FileViewHolder) convertView.getTag(); }else{ convertView = mInflater.inflate(R.layout.listitem2, null); holder = new FileViewHolder(convertView); convertView.setTag(holder); } holder.setData((File) data); } return convertView; }
此demo的github源码地址:
https://github.com/John-Chen/BlogSamples/tree/master/MultipleListTest
apk下载地址:
https://github.com/John-Chen/BlogSamples/blob/master/MultipleListTest/MultipleListTest.apk
如果写的有问题的地方,欢迎指教!