我正在使用 Hibernate 4,我在 JSF 页面中有一个过滤器来获取搜索结果。在执行搜索期间,我收到以下异常
java.lang.IllegalArgumentException: Parameter value [568903] did not match expected type [java.lang.Long] at org.hibernate.ejb.AbstractQueryImpl.validateParameterBinding(AbstractQueryImpl.java:370) at org.hibernate.ejb.AbstractQueryImpl.registerParameterBinding(AbstractQueryImpl.java:343) at org.hibernate.ejb.QueryImpl.setParameter(QueryImpl.java:370) at org.hibernate.ejb.QueryImpl.setParameter(QueryImpl.java:323)
下面是我的代码片段,我该如何解决这个问题?
private Long projectNo;
public Long getProjectNo() {
return projectNo;
}
public void setProjectNo(Long projectNo) {
this.projectNo = projectNo;
}
在 DAO 类中我有以下内容
String projectNo = filters.get("projectNo");
List<Predicate> criteria = new ArrayList<Predicate>();
if (projectNo!= null) {
ParameterExpression<String> pexp = cb.parameter(String.class, "projectNo");
Predicate predicate = cb.equal(emp.get(Project_.projectNo), pexp);
criteria.add(predicate);
}
TypedQuery<Project> q = entityManager.createQuery(c);
TypedQuery<Long> countquery = entityManager.createQuery(countQ);
q.setParameter("projectNo", projectNo); // error in this line
countquery.setParameter("projectNo", projectNo);
编辑 1
public void getProjects(ProjectQueryData data) {
在 ProjectQueryData
类中,我有以下构造函数
public ProjectQueryData (int start, int end, String field,
QuerySortOrder order, Map<String, String> filters) {
最佳答案
因为持久属性projectNo的类型是Long
,创建ParameterExpression时的类型参数应该是Long
。因此,因为 ParameterExpression 的类型是 Long
,所以参数值的类型也应该是 Long:
//because this persistent Attribute is Long:
private Long projectNo;
//we use Long here as well
ParameterExpression<Long> pexp = cb.parameter(Long.class, "projectNo");
...
//and finally set parameter. Long again, because that is the type
// type of ParameterExpression:
query.setParameter("projectNo", Long.valueOf(projectNo));
关于java - Hibernate 参数值 [568903] 与预期类型不匹配 [java.lang.Long],我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15229810/