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) {

Mikko Maunu · Accepted Answer · 2013-03-05 19:00:40Z

19

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));

answered Mar 5, 2013 at 19:00

Mikko Maunu

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2 Comments

Jacob Over a year ago

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

Philip Tenn Over a year ago

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

Philip Tenn · Accepted Answer · 2013-03-05 17:12:44Z

4

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.

answered Mar 5, 2013 at 17:12

Philip Tenn

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1 Comment

milosmns Over a year ago

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.

Mr. Polywhirl · Accepted Answer · 2014-12-30 11:15:22Z

3

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");

edited Dec 30, 2014 at 11:15

Mr. Polywhirl

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answered Mar 5, 2013 at 17:08

PSR

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7 Comments

Jacob Over a year ago

The following error

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

ninnemannk Over a year ago

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

Jacob Over a year ago

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

Jacob Over a year ago

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

Jacob Over a year ago

@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) {

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