假如一个学生类的list集合，返回50条数据；我只需要该集合中的学生姓名name字段，其他的字段都不想要，那么可以

String gender="男";
List<Student> students = studentMapper.selectListByQuery(QueryWrapper.create()
.where(STUDENT.GENDER.eq(gender)));
List<String> nameList = students.stream().map(Student::getName).collect(Collectors.toList());
//nameList存的全是name字段的值

通过stram流把Student集合转换成StudentAo集合，案例

 public R<Page<StudentAo>> sysGetPage(@RequestBody StudentVo vo) {Page<Student> page = studentService.page(vo.getPage(), vo.giveQuerywrapper());Page<StudentAo> pageAo =new Page<>();List<Student> records = page.getRecords();List<StudentAo> collect = records.stream().map(student -> {StudentAo ao = new StudentAo();BeanUtils.copyProperties(student, ao);return ao;}).collect(Collectors.toList());pageAo.setRecords(collect);pageAo.setPageSize(page.getPageSize());pageAo.setPageNumber(page.getPageNumber());pageAo.setTotalPage(page.getTotalPage());pageAo.setTotalRow(page.getTotalRow());return R.ok("成功",pageAo);}