Saven in table using JPA in STS

[eladerezador]

I´m trying to save an entry from a form using JPA+Hibernat, but the following error message comes up:

GRAVE: El Servlet.service() para el servlet [appServlet] en el contexto con ruta [/GoodProject] lanzó la excepción [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: org.hibernate.PropertyAccessException: could not set a field value by reflection setter of com.duckranger.goodproject.domain.Tramp.id; nested exception is javax.persistence.PersistenceException: org.hibernate.PropertyAccessException: could not set a field value by reflection setter of com.duckranger.goodproject.domain.Tramp.id] con causa raíz
java.lang.IllegalArgumentException: Can not set java.lang.Long field com.duckranger.goodproject.domain.Tramp.id to org.hibernate.id.IdentifierGeneratorHelper$2

This is the method Controller:

@RequestMapping(value = "/tramps", method = RequestMethod.POST, params="po4")
public String saveJPA(Model model, HttpServletRequest request) {
logger.info("Saving... Tramp in JPA");
Tramp tramp = new Tramp();
tramp.setName(request.getParameter("name"));
tramp.setDistance(12.4);
tramp.setDifficulty(null);
//model.addAttribute("tramps", trampsService.save(tramp));
tramp = trampsService.save(tramp);
return "tramps/add";
}

Which is the problem?

Thank you very much for your help

[eladerezador]

This is my service classes and the method save

public Tramp save(Tramp t) {
return trampRepository.save(t);
}

[Martin Lippert]

Hey!

I thought I already moved this from the STS group to another one, but it looks like I was wrong. So I moved it now into the Data forum since I guess it is a better fit. The question doesn't seem to be tooling related.

HTH,
Martin