No Dialect mapping for JDBC type: 1111

Question

I'm working on a Spring JPA Application, using MySQL as database. I ensured that all spring-jpa libraries, hibernate and mysql-connector-java is loaded.

I'm running a mysql 5 instance. Here is a excerpt of my application.properties file:

spring.jpa.show-sql=false
spring.jpa.hibernate.ddl-auto=create-drop
spring.jpa.database-platform=org.hibernate.dialect.MySQL5Dialect

spring.datasource.url=jdbc:mysql://localhost/mydatabase
spring.datasource.username=myuser
spring.datasource.password=SUPERSECRET
spring.datasource.driverClassName=com.mysql.jdbc.Driver

When executing an integration test, spring startsup properly but fails on creating the hibernate SessionFactory, with the exception:

org.hibernate.MappingException: No Dialect mapping for JDBC type: 1111

I think my dialects should be Mysql5Dialect, I also tried the one explicitly stating InnoDB, and the two dialect options which don't indicate the version 5. But I always end up with the same 'No Dialect mapping for JDBC type: 1111' message. My application.properties file resides in the test/resources source folder. It is recognized by the JUnit Test runner (I previously got an exception because of an typo in it).

Are the properties I'm setting wrong? I couldn't find some official documentation on these property names but found a hint in this stackoverflow answer: https://stackoverflow.com/a/25941616/1735497

Looking forward for your answers, thanks!

BTW The application is already using spring boot.

Answer 1

I got the same error because my query returned a UUID column. To fix that I returned the UUID column as varchar type through the query like "cast(columnName as varchar)", then it worked.

Example:

public interface StudRepository extends JpaRepository<Mark, UUID> {

    @Modifying
    @Query(value = "SELECT Cast(stuid as varchar) id, SUM(marks) as marks FROM studs where group by stuid", nativeQuery = true)
    List<Student> findMarkGroupByStuid();

    public static interface Student(){
        private String getId();
        private String getMarks();
    }
}

Answer 2

Here the answer based on the comment from SubOptimal:

The error message actually says that one column type cannot be mapped to a database type by hibernate. In my case it was the java.util.UUID type I use as primary key in some of my entities. Just apply the annotation @Type(type="uuid-char") (for postgres @Type(type="pg-uuid"))

Answer 3

There is also another common use-case throwing this exception. Calling function which returns void. For more info and solution go here.

Answer 4

Please Check if some Column return many have unknow Type in Query .

eg : '1' as column_name can have type unknown

and 1 as column_name is Integer is correct One .

This thing worked for me.

Answer 5

I got the same error, the problem here is UUID stored in DB is not converting to object.

I tried applying these annotations @Type(type="uuid-char") (for postgres @Type(type="pg-uuid") but it didn't work for me.

This worked for me. Suppose you want id and name from a table with a native query in JPA. Create one entity class like 'User' with fields id and name and then try converting object[] to entity we want. Here this matched data is list of array of object we are getting from query.

@Query( value = "SELECT CAST(id as varchar) id, name from users ", nativeQuery = true)

public List<Object[]> search();

public class User{
   private UUID id;
   private String name;
}


List<User> userList=new ArrayList<>();

for(Object[] data:matchedData){
        userList.add(new User(UUID.fromString(String.valueOf(data[0])),
                String.valueOf(data[1])));

    }

Suppose this is the entity we have

Answer 6

Finding the column that triggered the issue

First, you didn't provide the entity mapping so that we could tell what column generated this problem. For instance, it could be a UUID or a JSON column.

Now, you are using a very old Hibernate Dialect. The MySQL5Dialect is meant for MySQL 5. Most likely you are using a newer MySQL version.

So, try to use the MySQL8Dialect instead:

spring.jpa.database-platform=org.hibernate.dialect.MySQL8Dialect

Adding non-standard types

In case you got the issue because you are using a JSON column type, try to provide a custom Hibernate Dialect that supports the non-standard Type:

public class MySQL8JsonDialect
        extends MySQL8Dialect{
 
    public MySQL8JsonDialect() {
        super();
        this.registerHibernateType(
            Types.OTHER, JsonStringType.class.getName()
        );
    }
}

Ans use the custom Hibernate Dialect:

<property
    name="hibernate.dialect"
    value="com.vladmihalcea.book.hpjp.hibernate.type.json.MySQL8JsonDialect"
/>

If you get this exception when executing SQL native queries, then you need to pass the type via addScalar:

JsonNode properties = (JsonNode) entityManager
.createNativeQuery(
    "SELECT properties " +
    "FROM book " +
    "WHERE isbn = :isbn")
.setParameter("isbn", "978-9730228236")
.unwrap(org.hibernate.query.NativeQuery.class)
.addScalar("properties", JsonStringType.INSTANCE)
.getSingleResult();
 
assertEquals(
    "High-Performance Java Persistence",
    properties.get("title").asText()
);

Answer 7

Sometimes when you call sql procedure/function it might be required to return something. You can try returning void: RETURN; or string (this one worked for me): RETURN 'OK'

Answer 8

If you have native SQL query then fix it by adding a cast to the query.

Example:

CAST('yourString' AS varchar(50)) as anyColumnName

In my case it worked for me.

Answer 9

In my case the problem was that, I forgot to add resultClasses attribute when I setup my stored procedure in my User class.

@NamedStoredProcedureQuery(name = "find_email",
                procedureName = "find_email", resultClasses = User.class, //<--I forgot that. 
                parameters = {
                    @StoredProcedureParameter(mode = ParameterMode.IN, name = "param_email", type = String.class)
                }),

Answer 10

This also happens when you are using Hibernate and returning a void function. AT least w/ postgres. It doesnt know how to handle the void. I ended up having to change my void to a return int.

Answer 11

If you are using Postgres, check that you don't have a column of type Abstime. Abstime is an internal Postgres datatype not recognized by JPA. In this case, converting to Text using TO_CHAR could help if permitted by your business requirements.

Answer 12

Another simple explanation might be that you're fetching a complex Type (Entity/POJO) but do not specify the Entity to map to:

String sql = "select yourentity.* from {h-schema}Yourentity yourentity";
return entityManager.createNativeQuery(sql).getResultList();

simply add the class to map to in the createNativeQuery method:

return entityManager.createNativeQuery(sql, Yourentity.class).getResultList();

Answer 13

For anybody getting this error with an old hibernate (3.x) version:

do not write the return type in capital letters. hibernate type implementation mapping uses lowercase return types and does not convert them:

CREATE OR REPLACE FUNCTION do_something(param varchar)
    RETURNS integer AS
$BODY$
...

Answer 14

This is for Hibernate (5.x) version

Calling database function which return JSON string/object

For this use unwrap(org.hibernate.query.NativeQuery.class).addScalar() methods for the same.

Example as below (Spring & Hibernate):

@PersistenceContext

EntityManager em;

@Override

    public String getJson(String strLayerName) {

        String *nativeQuery* = "select fn_layer_attributes(:layername)";

        return em.createNativeQuery(*nativeQuery*).setParameter("layername", strLayerName).**unwrap(org.hibernate.query.NativeQuery.class).addScalar**("fn_layer_attributes", **new JsonNodeBinaryType()**) .getSingleResult().toString();

    }

Answer 15

Function or procedure returning void cause some issue with JPA/Hibernate, so changing it with return integer and calling return 1 at the end of procedure may solved the problem.

SQL Type 1111 represents String.

Answer 16

In my case, the issue was Hibernate not knowing how to deal with an UUID column. If you are using Postgres, try adding this to your resources/application.properties:

spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.PostgreSQL9Dialect

Answer 17

if using Postgres

public class CustomPostgreSqlDialect extends PostgreSQL94Dialect{

    @Override
    public SqlTypeDescriptor remapSqlTypeDescriptor(SqlTypeDescriptor sqlTypeDescriptor)
    {
        switch (sqlTypeDescriptor.getSqlType())
        {
        case Types.CLOB:
            return VarcharTypeDescriptor.INSTANCE;
        case Types.BLOB:
            return VarcharTypeDescriptor.INSTANCE;
        case 1111://1111 should be json of pgsql
            return VarcharTypeDescriptor.INSTANCE;
        }
        return super.remapSqlTypeDescriptor(sqlTypeDescriptor);
    }
    public CustomPostgreSqlDialect() {
        super();
        registerHibernateType(1111, "string");
    }}

and use

<prop key="hibernate.dialect">com.abc.CustomPostgreSqlDialect</prop>

Answer 18

In my case it was solved by adding the columnDefinition:

@Column(columnDefinition = "json")