通过将标准与android中的改进进行比较，筛选JSON对象列表

这是我显示学生名单的代码

private void requestData(String uri) {

    RestAdapter adapter = new RestAdapter.Builder()
        .setEndpoint(ENDPOINT)
        .build();

    StudentsAPI api = adapter.create(StudentsAPI.class);

    api.getFeed(new Callback<List<Student>>() {

        @Override
        public void success(List<Student> arg0, Response arg1) {
            studentList = arg0;
            updateDisplay();
        }

        @Override
        public void failure(RetrofitError arg0) {
            // TODO Auto-generated method stub

        }
      });

//接口

public interface StudentsAPI {

@GET("/data/students.json")
public void getFeed(Callback<List<Student>> response);

}

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我想通过比较课程来筛选列表。如何比较字符串并显示结果

我希望您必须使用ViewListView、RecyclerView、GridView来显示数据，因此，可以根据studentList列表中的元素和类似位置，在List/Recycler/GridView适配器的getView或onBindViewHolder方法中写入比较逻辑

 @Override
public CustomViewHolder onCreateViewHolder(ViewGroup viewGroup, int i) {
    View view = LayoutInflater.from(viewGroup.getContext()).inflate(R.layout.student_row, null);

    CustomViewHolder viewHolder = new CustomViewHolder(view);
    return viewHolder;
}
@Override
public void onBindViewHolder(CustomViewHolder customViewHolder, int i) {
    Student student = studentList.get(i);

    if(student.getCourse()==filter){
                customViewHolder.textView.setText(student.getName());

    }
    else{
         customViewHolder.getView().setVisibility(View.Gone);
    }


}