CODEWITHSUNDEEP
javahibernatejpajpa-2.0
Jacob
Jacob

Reputation: 14731

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) {

Upvotes: 17

Views: 116995

Answers (3)

PSR
PSR

Reputation: 40318

projectNo is long type in DAO, so change it to long.

Try this:

q.setParameter("projectNo", new Long(projectNo));

I think you should change:

ParameterExpression<String> pexp = cb.parameter(Long.class, "projectNo");

to

ParameterExpression<String> pexp = cb.parameter(String.class, "projectNo");

Upvotes: 3

Mikko Maunu
Mikko Maunu

Reputation: 42084

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));

Upvotes: 19

Philip Tenn
Philip Tenn

Reputation: 6073

In your DAO class, you are getting projectNo as a String:

String projectNo = filters.get("projectNo");

However, in your model class, you are defining projectNo as a Long.

When you set parameter in your DAO on this line: 

q.setParameter("projectNo", projectNo); // error in this line

You are setting the parameter as a String. Try changing that line as follows (assuming that you've null-checked projectNo):

q.setParameter("projectNo", Long.parseLong(projectNo));

It also probably wouldn't hurt (defensive programming) to be sure that projectNo is numeric prior to calling Long.parseLong. You can do this with Apache Commons StringUtils.isNumeric.

Upvotes: 3

Related Questions