Spring supports creating queries by examples of the object to look for. Like:
//if not setting the age, it will always look for age=0 Person p = new Person(); p.setLastName("Smith"); List<Person> foundPersons = personRepository.findAll(Example.of(p)); @Entity public class Person { private String firstName; private String lastName; private LocalDate dob; private int age; }
Problem: if the @Entity
has primitive fields, then their default value will actually be used for creating the query. The example above will result in:
SELECT * from persons where lastname := 'Smith' and age := 0
In my example I have a database field where age
must always be filled, thus is not allowed to be null
. Therefore the entity has a primitive int age
field.
Of course I could now change the field to Integer age
, but then I’d marking the field being an optional nullable attribute, which is not true.
So, how can I skip primitives that have not been set on the Example
?
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Answer
Yes, you can do it:
Person p = new Person(); p.setLastName("Smith"); Example criteria = Example.create(p).setPropertySelector( Example.NotNullOrZeroPropertySelector.INSTANCE ); List<Person> foundPersons = session.createCriteria(Person.class).add(criteria).list();
Example.NotNullOrZeroPropertySelector.INSTANCE
is a property selector that includes only properties that are not null
and non-zero (if numeric)
UPD
Above an example for Hibernate org.hibernate.criterion.Example
class. For org.springframework.data.domain.Example
you can ignore primitive fields by manually specifying names of these fields:
Person p = new Person(); p.setLastName("Smith"); ExampleMatcher matcher = ExampleMatcher.matching().withIgnorePaths("age").withIgnoreNullValues(); Example criteria = Example.of(p, matcher); List<Person> foundPersons = personRepository.findAll(criteria);