JPA/Criteria API - Like & equal problem

Question

I'm trying to use Criteria API in my new project:

public List<Employee> findEmps(String name) {
    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
    Root<Employee> emp = c.from(Employee.class);
    c.select(emp);
    c.distinct(emp);
    List<Predicate> criteria = new ArrayList<Predicate>();

    if (name != null) {
        ParameterExpression<String> p = cb.parameter(String.class, "name");
        criteria.add(cb.equal(emp.get("name"), p));
    }

    /* ... */

    if (criteria.size() == 0) {
        throw new RuntimeException("no criteria");
    } else if (criteria.size() == 1) {
        c.where(criteria.get(0));
    } else {
        c.where(cb.and(criteria.toArray(new Predicate[0])));
    }

    TypedQuery<Employee> q = em.createQuery(c);

    if (name != null) {
        q.setParameter("name", name);
    }

    /* ... */

    return q.getResultList();
}

Now when I change this line:

            criteria.add(cb.equal(emp.get("name"), p));

to:

            criteria.add(cb.like(emp.get("name"), p));

I get an error saying:

The method like(Expression, Expression) in the type CriteriaBuilder is not > applicable for the arguments (Path, ParameterExpression)

What's the problem?

axtavt · Accepted Answer · 2010-10-25T12:26:35.960

80

Perhaps you need

criteria.add(cb.like(emp.<String>get("name"), p));

because first argument of like() is Expression<String>, not Expression<?> as in equal().

Another approach is to enable generation of the static metamodel (see docs of your JPA implementation) and use typesafe Criteria API:

criteria.add(cb.like(emp.get(Employee_.name), p));

(Note that you can't get static metamodel from em.getMetamodel(), you need to generate it by external tools).

edited Oct 25 '10 at 12:26

answered Oct 25 '10 at 12:18

axtavt

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"name" here is the variable name of the parameter here? like :name in jpql? – Kalpesh Soni Mar 28 '14 at 21:09
@axtavt, how i can ignore case – Black Swan Jan 22 '18 at 03:26
6

will the result of the query be like `employee.name like '%%'`? Is it possible to achieve `like '%'`? – Mohammad Faisal Apr 16 '18 at 11:55
Of course, in `builder.like()` method, the second arg can be a String. So you can put sth like `Predicate like = builder.like(root.get(Employee_.name), "%joke%")`. – WesternGun Sep 08 '21 at 14:03

score 24 · Answer 2 · edited Jul 25 '16 at 09:47

24

Better: predicate (not ParameterExpression), like this :

List<Predicate> predicates = new ArrayList<Predicate>();
if(reference!=null){
    Predicate condition = builder.like(root.<String>get("reference"),"%"+reference+"%");
    predicates.add(condition);
}

edited Jul 25 '16 at 09:47

Tunaki

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answered Jan 31 '16 at 18:40

user3077341

239
2
5

Also, you can add a replace in there to make it easier to use, for example: Predicate condition = builder.like(root.get("reference"),reference.replace("*", "%")); This way, you can pass on * as special characters to represent the like behavior. – Paulo 'PaulusHC' Cruz Sep 01 '18 at 13:11

score 4 · Answer 3 · answered Jan 08 '21 at 17:10

4

It will work with a small addition of .as(String.class):

criteria.add(cb.like(emp.get("name").as(String.class), p));

answered Jan 08 '21 at 17:10

Christoph

194
1
8

score 2 · Answer 4 · answered Sep 21 '18 at 14:52

2

Use :

personCriteriaQuery.where(criteriaBuilder.like(
criteriaBuilder.upper(personRoot.get(Person_.description)), 
"%"+filter.getDescription().toUpperCase()+"%"));

answered Sep 21 '18 at 14:52

Abdessamad HALLAL

121
1
8

score 1 · Answer 5 · answered Sep 30 '20 at 20:37

I tried all the proposals and although it did not give an error, it did not recover anything if I did not put all the expected text, in the end what was proposed by user3077341 worked for me but with the variant that I use '%' instead of "%", the like worked perfect.

List<Predicate> predicates = new ArrayList<Predicate>();
if(reference!=null){
    Predicate condition = builder.like(root.<String>get("reference"),'%'+reference+'%');
    predicates.add(condition);
}

You mean a char instead of a String? Oh. – WesternGun Sep 08 '21 at 14:04 — WesternGun, Sep 08 '21 at 14:04

JPA/Criteria API - Like & equal problem

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