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(用户, 可分页, users.size());
但现在我在排序和分页本身方面遇到了问题。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);
}
这是我的用户仓库
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);
我也有同样的问题。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());
如参考文档中所示,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,以确保它们仅读取请求的 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 Pageable
对象中定义。无需获取列表,然后尝试从中创建页面。
您应该按照 dubonzi 的answer的建议进行操作。
如果您仍然想对给定的
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
如果需要排序,请使用另一个 PagedListHolder 构造函数和 MutableSortDefinition。
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());
}
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,我们也可以返回实体的不同页面。
在 JHipster 框架中有一个用于此类操作的接口
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());
}
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());
您可以使用此通用函数将列表转换为页面。
public static<T> Page<T> convertToPage(List<T> objectList, Pageable pageable){
int start = (int) pageable.getOffset();
int end = Math.min(start+pageable.getPageSize(),objectList.size());
List<T> subList = start>=end?new ArrayList<>():objectList.subList(start,end);
return new PageImpl<>(subList,pageable,objectList.size());
}
//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;
}
}
你没有做出分页结果
new PageImpl<User>(users, pageable, users.size());
不会隐式生成分页结果,
在这种情况下,
pageable
参数只是创建Page对象的元数据,如页面,偏移量,大小...等
所以你必须使用像这样的存储库方法
Page<User>findAllByProgramId(Integer programId, Pageable pageable);
谢谢大家,下面的代码适用于我的情况
int start = pageble.getOffset();
int end = (start + pageble.getPageSize()) > vehicleModelsList.size() ? vehicleModelsList.size() : (start + pageble.getPageSize());
您是否尝试过将存储库扩展到
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:
不要返回完整的数组列表,而是根据您的要求使用子列表。 您将从请求正文中的“可分页”对象中获取“偏移量”和大小。
new PageImpl<User>(users.subList(start, end), pageable, users.size());
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);
}
我也面临着同样的问题。我通过先获取总值,然后设置PageImpl
解决了这个问题Long total = users.size();
Page<User> pages = new PageImpl<User>(users, pageable, total );
从 SpringBoot 2.4 开始,您可以使用
PageableExecutionUtils
,如下所示:
Query query = new Query();
query.addCriteria(Criteria.where("ownedParkByOwner").ne(true));
List<PlaceV2> placeV2s = mongoTemplate.find(query, PlaceV2.class);
Page<PlaceV2> placeV2Page = PageableExecutionUtils.getPage(
placeV2s,
pageable,
placeV2s::size);