Spring中List到Page的转换

Question

我正在尝试在春季将列表转换为页面。我已经使用

对其进行了转换

new PageImpl(users, pageable, users.size());

但现在我遇到了排序和分页本身的问题。当我尝试传递大小和页面时，分页不起作用。

这是我正在使用的代码。

我的控制器

    public ResponseEntity<User> getUsersByProgramId(
        @RequestParam(name = "programId", required = true) Integer programId Pageable pageable) {

    List<User> users = userService.findAllByProgramId(programId);
    Page<User> pages = new PageImpl<User>(users, pageable, users.size());

    return new ResponseEntity<>(pages, HttpStatus.OK);
}

这是我的用户回购

public interface UserRepo extends JpaRepository<User, Integer>{

public List<User> findAllByProgramId(Integer programId);

这是我的服务

    public List<User> findAllByProgramId(Integer programId);

Answer 1

这是分页列表的正确答案

public ResponseEntity<User> getUsersByProgramId(
            @RequestParam(name = "programId", required = true) Integer programId, Pageable pageable) {
        List<User> users = userService.findAllByProgramId(programId);

        final int toIndex = Math.min((pageable.getPageNumber() + 1) * pageable.getPageSize(),
                    bidList.size());
        final int fromIndex = Math.max(toIndex - pageable.getPageSize(), 0);
        Page<User> pages = new PageImpl<User>(users.subList(fromIndex, toIndex), pageable, users.size());
    
        return new ResponseEntity<>(pages, HttpStatus.OK);
}

Answer 2

我遇到了同样的问题。我使用了子列表：

final int start = (int)pageable.getOffset();
final int end = Math.min((start + pageable.getPageSize()), users.size());
final Page<User> page = new PageImpl<>(users.subList(start, end), pageable, users.size());

Answer 3

在JHipster框架中有一个接口PageUtil：

static <T> Page<T> createPageFromList(List<T> list, Pageable pageable) {
    if (list == null) {
        throw new IllegalArgumentException("To create a Page, the list mustn't be null!");
    }

    int startOfPage = pageable.getPageNumber() * pageable.getPageSize();
    if (startOfPage > list.size()) {
        return new PageImpl<>(new ArrayList<>(), pageable, 0);
    }

    int endOfPage = Math.min(startOfPage + pageable.getPageSize(), list.size());
    return new PageImpl<>(list.subList(startOfPage, endOfPage), pageable, list.size());
}

Answer 4

基于@shilaimuslm 评论实现。在这种情况下，如果 start > end 在 subList 中，则不会抛出异常。

List<User> users = // ...

Pageable paging = PageRequest.of(pagePagination, sizePagination);
int start = Math.min((int)paging.getOffset(), users.size());
int end = Math.min((start + paging.getPageSize()), users.size());

Page<User> page = new PageImpl<>(users.subList(start, end), paging, users.size());

Answer 5

根据您的要求，不要返回完整的数组列表，而是取 subblist。 您将从请求正文中的“可分页”对象中获得“偏移”和大小。

new PageImpl<User>(users.subList(start, end), pageable, users.size());

Answer 6

你应该按照 dubonzi 的 answer 的建议去做。

如果您仍想对给定的List 使用分页，请使用PagedListHolder：

List<String> list = // ...

// Creation
PagedListHolder page = new PagedListHolder(list);
page.setPageSize(10); // number of items per page
page.setPage(0);      // set to first page

// Retrieval
page.getPageCount(); // number of pages 
page.getPageList();  // a List which represents the current page

如果您需要排序，请使用另一个 PagedListHolder constructor 和 MutableSortDefinition。

Answer 7

这可能是解决方案。排序和分页也可以这样工作：

控制器：

public ResponseEntity<User> getUsersByProgramId(
   @RequestParam(name = "programId", required = true) Integer programId Pageable pageable) {
       Page<User> usersPage = userService.findAllByProgramId(programId, pageable);
       Page<User> pages = new PageImpl<User>(usersPage.getContent(), pageable, usersPage.getTotalElements());

       return new ResponseEntity<>(pages, HttpStatus.OK);
}

服务：

Page<User> findAllByProgramId(Integer programId, Pageable pageable);

存储库：

public interface UserRepo extends JpaRepository<User, Integer>{
       public Page<User> findAllByProgramId(Integer programId, Pageable pageable);
}

这样，我们也可以返回不同的实体页面。

Answer 8

为此有一个Page implementation：

Page<Something> page = new PageImpl<>(yourList);

Answer 9

Try This:

public Page<Patient> searchPatientPage(SearchPatientDto patient, int page, int size){
        List<Patient> patientsList = new ArrayList<Patient>();
        Set<Patient> list=searchPatient(patient);
        patientsList.addAll(list);
        int start =  new PageRequest(page, size).getOffset();
        int end = (start + new PageRequest(page, size).getPageSize()) > patientsList.size() ? patientsList.size() : (start + new PageRequest(page, size).getPageSize());

        return new PageImpl<Patient>(patientsList.subList(start, end), new PageRequest(page, size), patientsList.size());
    }

Answer 10

//1) For a boot application create a paging repository interface
public interface PersonRepository extends PagingAndSortingRepository<Person, 
String> {
// Common CURD method are automatically implemented
}
//2) create a service Interface
public interface PersonService {
    Page<Person> listAllByPage(Pageable pageable); // Use common CURD findAll() method
    Page<Object> listSpecByPage(Pageable pageable, String x);

}
//3) create a service Impl Class of service interface
@Service
public class PersonServiceImpl implements PersonService {

    final PersonRepository personRepository;

    @Autowired
    PersonServiceImpl(PersonRepository personRepository){
    this.personRepository = personRepository;
    }

    @Override
    public Page<Person> listAllByPage(Pageable pageable) {
          return personRepository.findAll(pageable);
    }
    @Override
    public Page<Object> listSpecByPage(Pageable pageable, String path) {
        List<Object> objectlist = new ArrayList<Object>();
        // Do your process to get output in a list by using node.js run on a *js file defined in 'path' varriable
        Page<Object> pages1 = new PageImpl<Object>(objectlist, pageable, objectlist.size());
        return pages1;
    }

}
//4) write your controller
public class PersonController {

    final PersonService personService;

    @Autowired
    PersonController( PersonService personService ){
        this.personService = personService;
    }

    @GetMapping("/get") // Use of findALL() function
    Page<Person> listed( Pageable pageable){
        Page<Person> persons = personService.listAllByPage(pageable);
        return persons;
    } 
    @GetMapping("/spec") // Use of defined function
    Page<Object> listSpec( Pageable pageable, String path){
        Page<Object> obj = personService.listSpecByPage(pageable, path);
        return obj;
   } 

}

Answer 11

您是否尝试将存储库扩展到PagingAndSortingRepository？

public interface UserRepo extends PagingAndSortingRepository<Ticket, Integer> {
    Page<User> findAllByProgramId(Integer programId, Pageable pageable);
}

服务

Page<User> findAllByProgramId(Integer programId, Pageable pageable);

我假设您正在使用服务接口：

Answer 12

感谢下面的代码在我的情况下工作

    int start = pageble.getOffset();
    int end = (start + pageble.getPageSize()) > vehicleModelsList.size() ? vehicleModelsList.size() : (start + pageble.getPageSize());

Answer 13

如reference documentation 中所述，Spring Data 存储库通过简单地声明Pageable 类型的参数来支持查询方法的分页，以确保它们只读取请求的Page 所需的数据。

Page<User> page = findAllByProgramId(Integer programId, Pageable pageable);

这将返回一个Page 对象，其页面大小/设置在您的Pageable 对象中定义。无需获取列表然后尝试从中创建页面。