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Hibernate Parameter value [568903] did not match expected type [java.lang.Long]

I am using Hibernate 4 and I have a filter in JSF page to get search results. During execution of search I am getting the following exception

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)

Below is my code snippet, how can I fix this issue?

private Long projectNo; 

public Long getProjectNo() {
    return projectNo;
}

public void setProjectNo(Long projectNo) {
    this.projectNo = projectNo;
}

And in DAO class I have the following

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);

Edit 1

public void getProjects(ProjectQueryData data) { 

and in ProjectQueryData class, I have the following as constructor

public ProjectQueryData (int start, int end, String field,
            QuerySortOrder order, Map<String, String> filters) {
like image 831
Jacob Avatar asked Mar 05 '13 17:03

Jacob


1 Answers

Because type of persistent attribute projectNo is Long, type argument when creating ParameterExpression should be Long. And consequently, because type of the ParameterExpression is Long, type of the parameter's value should be Long as well:

//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));
like image 103
Mikko Maunu Avatar answered Oct 23 '22 05:10

Mikko Maunu