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

Question

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) {

Solution 1

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));

Solution 2

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"); 

Solution 3

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.

Comments

[Jacob]

The following error java.lang.IllegalArgumentException: Named parameter [projectNo] type mismatch; expecting [java.lang.String], found [java.lang.Long]

[ninnemannk]

Yout put <code>ParameterExpression<String> pexp = cb.parameter(String.class, "projectNo");</code> Change this to: ParameterExpression<String> pexp = cb.parameter(Long.class, "projectNo");

[Jacob]

@PSR Sorry I didn't include that. filter is Map<String, String> filters = data.getFilters();

[Jacob]

@PSR I have added that code as Edit 1 in my original question.

[Jacob]

@PSR I have the following method in a class and this class extends LazyDataModel public List<Project> load(int first, int pageSize, String sortField, SortOrder sortOrder, Map<String, String> filters) {

[Jacob]

let us continue this discussion in chat

[Jacob]

@PSR chat here

[Jacob]

Mikko, Thank you so much, this really has helped to solve the problem. Much appreciated.

[Philip Tenn]

+1 nice job, looks like you were able to piece everything together for a cohesive answer that worked!

[milosmns]

Adding megabytes of Apache libs to your project in order to get one util method isn't defensive, it's actually a bad idea. A simple try/catch works just fine in this case.