我已经写了公司课程的标准。
以下是公司类别,companySearch类别和条件。但是条件列表是抛出异常。例外是“ org.hibernate.QueryException:无法解析属性:com.sesami.common.domain.Company的san.san”。如何访问Company.san.san?
公司类
public class Company extends DomainObject implements UserDetails {
private Long id;
private String companyName;
private CompanyType companyType;
private String description;
private String companyURL;
private String billToEmail;
private String hashPassword;
private SAN san;
@OneToOne(cascade = { CascadeType.ALL })
public SAN getSan() {
return san;
}
public void setSan(SAN san) {
this.san = san;
}
...
}
公司搜索
public class CompanySearch {
private String companyName;
private String email;
private Long san;
private String gstNumber;
......
public Long getSan() {
return san;
}
public void setSan(Long san) {
this.san = san;
}
...
}
标准
companyCriteria = this.getSession().createCriteria(
Company.class);
if (companySearch.getSan() != null
&& !"".equals(companySearch.getSan()))
companyCriteria.add(Restrictions.eq("san.san",
companySearch.getSan()));
Integer count = ((Long) companyCriteria.setProjection(
Projections.rowCount()).uniqueResult()).intValue();
companyCriteria.setProjection(null);
companyCriteria.setResultTransformer(Criteria.ROOT_ENTITY);
companyCriteria
.setFirstResult((pager.getPage() - 1) * pager.getPageSize())
.setMaxResults(pager.getPageSize()).list();
List<Company> companies = companyCriteria.list();
PagedResultSet pr = new PagedResultSet();
pr.setPager(pager);
pr.setResultSet(companies);
pr.setRowCount(count);
return pr;
最佳答案
您必须使用子条件或别名创建San实体的联接:
companyCriteria.createAlias("san", "sanAlias");
companyCriteria.add(Restrictions.eq("sanAlias.san",
companySearch.getSan()));
要么
companyCriteria.createCriteria("san").add(Restrictions.eq("san",
companySearch.getSan()));
Hibernate reference documentation甚至Criteria javadoc中对此都有很好的解释。
注意,这与Spring完全无关,与Hibernate无关。如果您在Spring文档中搜索了如何执行此操作,那也就不足为奇了。