Error with JPA PagingAndSorting request parameter - spring

Question

I have a Controller that pages and sorts all the news in my database:
#RequestMapping(value = "/viewstatus", method = RequestMethod.GET)
ModelAndView viewStatus(ModelAndView modelAndView, #RequestParam(name = "p", defaultValue = "1") int pageNumber) {
Page<StatusUpdate> page = statusUpdateService.getPage(pageNumber);
modelAndView.getModel().put("page", page);
modelAndView.setViewName("app.viewStatus");
return modelAndView;
}
With its call to the service that works fine:
public Page<StatusUpdate> getPage(int pageNumber) {
PageRequest request = new PageRequest(pageNumber-1, pageSize, Sort.Direction.DESC, "added");
return statusUpdateDao.findAll(request);
}
But now, I would like to do the same SortingAndPaging BUT with one parameter (SiteUser). Here is my object:
#Entity
#Table(name = "status_update")
public class StatusUpdate {
#Id
#Column(name = "id")
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#Size(min=5, max=255, message="{addstatus.title.size}")
#Column(name = "title")
private String title;
#Size(min=5, max=5000, message="{addstatus.text.size}")
#Column(name = "text")
private String text;
#Column(name = "added")
#Temporal(TemporalType.TIMESTAMP)
#DateTimeFormat(pattern="yyyy/MM/dd hh:mm:ss")
private Date added;
#OneToOne(targetEntity = SiteUser.class)
#JoinColumn(name="user_id")
private SiteUser siteUser;
#PrePersist
protected void onCreate() {
if (added == null) {
added = new Date();
}
}
public StatusUpdate() {
}
But when I do it, it gives me this error:
Exception: org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [3] did not match expected type [com.caveofprogramming.model.entity.SiteUser (n/a)]; nested exception is java.lang.IllegalArgumentException: Parameter value [3] did not match expected type [com.caveofprogramming.model.entity.SiteUser (n/a)]
Failed URL: http://192.168.160.128:8080/viewmystatus
Exception message: Parameter value [3] did not match expected type [com.caveofprogramming.model.entity.SiteUser (n/a)]; nested exception is java.lang.IllegalArgumentException: Parameter value [3] did not match expected type [com.caveofprogramming.model.entity.SiteUser (n/a)]
I tired to send a String, but it did not work. I had to change it to a NativeQuery but I does not work with PagingAndSorting, it only works as a List (which is a headache to work with). So if you can help me with the ERROR of PagingAndSorting that would be great.
Here is the Controller:
#RequestMapping(value = "/viewmystatus", method = RequestMethod.GET)
ModelAndView viewMyStatus(ModelAndView modelAndView, #RequestParam(name = "p", defaultValue = "1") int pageNumber) {
SiteUser user = getUser();
Long user_id= user.getId();
Page<StatusUpdate> page = statusUpdateService.findMyStatusUpdates(user_id, pageNumber);
for(StatusUpdate statusUpdate: page){
SiteUser siteUser= statusUpdate.getSiteUser();
modelAndView.getModel().put("siteuser", siteUser);
}
modelAndView.getModel().put("page", page);
modelAndView.setViewName("app.viewStatus");
return modelAndView;
}
Here is the service:
public Page<StatusUpdate> findMyStatusUpdates(Long user_id, int pageNumber) {
PageRequest request = new PageRequest(pageNumber-1, pageSize, Sort.Direction.DESC, "added");
return statusUpdateDao.findBySiteUser(user_id, request);
}
And the DAO:
#Repository
public interface StatusUpdateDao extends PagingAndSortingRepository<StatusUpdate, Long> {
StatusUpdate findFirstByOrderByAddedDesc();
Page<StatusUpdate> findBySiteUser(Long user_id, Pageable pageable);
}
Thanks for your help!

Answer 1

Use a SiteUser object:
public Page<StatusUpdate> findMyStatusUpdates(Long user_id, int pageNumber) {
PageRequest request = new PageRequest(pageNumber-1, pageSize, Sort.Direction.DESC, "added");
return statusUpdateDao.findBySiteUser(new SiteUser(user_id), request);
}
You could try query by nested properties as well, but I think you must change user_id by userId, as underscore is a reserved character. Have a look to the documentation:
Spring data JPA Property expressions
Have you changed your findBySiteUser method? The one you posted looks fine but it seems is complaining about the signature. Even when PageRequest implements Pageable the signature has to be declared, explicitly, using Pageable, but the error message you are getting says PageRequest
public abstract org.springframework.data.domain.Page com.caveofprogramming.model.repository.StatusUpdateDao.findB‌​ySiteUser(com.caveof‌​programming.model.en‌​tity.SiteUser,org.sp‌​ringframework.data.d‌​omain.PageRequest)
More info:
PageRequest parameter not recognized as Pageable in Paging query