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.
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();
}
}
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"))
There is also another common use-case throwing this exception. Calling function which returns void. For more info and solution go here.
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
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.
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()
);
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'
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.
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
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();
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)
}),
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.
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.
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>
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
$BODY$
...
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();
}
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.
If you are calling EntityManager.createNativeQuery(), be sure to include the resulting java class in the second parameter:
return em.createNativeQuery(sql, MyRecord.class).getResultList()
After trying many proposed solutions, including:
https://stackoverflow.com/a/59754570/349169 which is one of the solutions proposed here
https://vladmihalcea.com/hibernate-no-dialect-mapping-for-jdbc-type/
it was finally this one that fixed everything with the least amount of changes:
https://gist.github.com/agrawald/adad25d28bf6c56a7e4618fe95ee5a39
The trick is to not have #TypeDef on your class, but instead have 2 different #TypeDef in 2 different package-info.java files. One inside your production code package for your production DB, and one inside your test package for your test H2 DB.
I am working with Spring Data 2.0.6.RELEASE.
I am working about pagination for performance and presentation purposes.
Here about performance I am talking about that if we have a lot of records is better show them through pages
I have the following and works fine:
interface PersonaDataJpaCrudRepository extends PagingAndSortingRepository<Persona, String> {
}
The #Controller works fine with:
#GetMapping(produces=MediaType.TEXT_HTML_VALUE)
public String findAll(Pageable pageable, Model model){
Through Thymeleaf I am able to apply pagination. Therefore until here the goal has been accomplished.
Note: The Persona class is annotated with JPA (#Entity, Id, etc)
Now I am concerned about the following: even when pagination works in Spring Data about the amount the records, what about of the content of each record?.
I mean: let's assume that Persona class contains 20 fields (consider any entity you want for your app), thus for a view based in html where a report only uses 4 fields (id, firstname, lastname, date), thus we have 16 unnecessary fields for each entity in memory
I have tried the following:
interface PersonaDataJpaCrudRepository extends PagingAndSortingRepository<Persona, String> {
#Query("SELECT p.id, id.nombre, id.apellido, id.fecha FROM Persona p")
#Override
Page<Persona> findAll(Pageable pageable);
}
If I do a simple print in the #Controller it fails about the following:
java.lang.ClassCastException:
[Ljava.lang.Object; cannot be cast to com.manuel.jordan.domain.Persona
If I avoid that the view fails with:
Caused by:
org.springframework.expression.spel.SpelEvaluationException:
EL1008E:
Property or field 'id' cannot be found on object of type
'java.lang.Object[]' - maybe not public or not valid?
I have read many posts in SO such as:
java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to
I understand the answer and I am agree about the Object[] return type because I am working with specific set of fields.
Is mandatory work with the complete set of fields for each entity? Should I simply accept the cost of memory about the 16 fields in this case that never are used? It for each record retrieved?
Is there a solution to work around with a specific set of fields or Object[] with the current API of Spring Data?
Have a look at Spring data Projections. For example, interface-based projections may be used to expose certain attributes through specific getter methods.
Interface:
interface PersonaSubset {
long getId();
String getNombre();
String getApellido();
String getFecha();
}
Repository method:
Page<PersonaSubset> findAll(Pageable pageable);
If you only want to read a specific set of columns you don't need to fetch the whole entity. Create a class containing requested columns - for example:
public class PersonBasicData {
private String firstName;
private String lastName;
public PersonBasicData(String firstName, String lastName) {
this.firstName = fistName;
this.lastName = lastName;
}
// getters and setters if needed
}
Then you can specify query using #Query annotation on repository method using constructor expression like this:
#Query("SELECT NEW some.package.PersonBasicData(p.firstName, p.lastName) FROM Person AS p")
You could also use Criteria API to get it done programatically:
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<PersonBasicData> query = cb.createQuery(PersonBasicData.class);
Root<Person> person = query.from(Person.class);
query.multiselect(person.get("firstName"), person.get("lastName"));
List<PersonBasicData> results = entityManager.createQuery(query).getResultList();
Be aware that instance of PersonBasicData being created just for read purposes - you won't be able to make changes to it and persist those back in your database as the class is not marked as entity and thus your JPA provider will not work with it.
I have a spring data jpa repository. I want to query a database view.
The database is DB2 and supports a feature called 'global variable' which actually is a session variable. The db view declaration uses a global variable.
How do I set the value of this variable at runtime before the view is executed?
Would something like this work?
public interface DomainRepository extends org.springframework.data.jpa.repository.JpaRepository<Domain, IdType> {
#Query(value = "SET SCHEMA.VAR_GLOBAL = :param; SELECT * FROM SCHEMA.DOMAIN", nativeQuery = true)
List<Domain> findByDomain(#Param("param") String param);
}
Are there any alternative solutions?
That's what I want to achieve in SQL:
SET CURRENT APPLICATION COMPATIBILITY = 'V11R1';
SET SCHEMA.VAR_GL = 'Value';
SELECT * FROM SCHEMA.VIEW;
The SCHEMA.VIEW is declared as:
SELECT * FROM SCHEMA.TABLE WHERE field = VAR_GL
I have to mark the public method on the controller with #Transactional annotation to execute the queries together in one db session.
public interface Controller {
#Transactional
List<Options> loadOptions();
}
public class ControllerImpl implements Controller {
#Autowired
private DomainRepository repo;
#Override
public List<Option> loadOptions() {
this.repo.setCompatibilityMode();
this.repo.setGlobalVariableA("Value");
List<Option> list = this.repo.loadDropdown();
return list;
}
In my jpa repository I need a method for each global variable:
public interface DomainRepository extends JpaCrudRepository<Option, OptionPK> {
#Modifying
#Query(value = "SET CURRENT APPLICATION COMPATIBILITY = 'V11R1'", nativeQuery = true)
void setCompatibilityMode();
}
#Modifying
#Query(value = "set SCHEMA.VAR_GL = :value", nativeQuery = true)
void setGlobalVariableA(#Param("value") String value);
#Query(value = "SELECT * FROM SCHEMA.VIEW", nativeQuery = true)
List<Option> loadDropdown();
}
That's how it works. I'd like to improve this solution further but I have no idea how to make the variable query more reusable by making the variable name a parameter.
Variable declarations don't work over JDBC but temp tables do:
DECLARE GLOBAL TEMPORARY TABLE SESSION.myvars
(
VAR_GLOBAL VARCHAR(255)
) ON COMMIT PRESERVE ROWS;
You cannot lump two statements together like in your example, but you should be able to create two different Querys and execute them in separate calls to executeUpdate() and getResultList() respectively. (I'm not very familiar with JPA, I may have used wrong method names, hopefully you get the idea.)
You can try CRUD methods:
public interface DomainRepository extends CrudRepository<Domain, Long>{
List<Domain> findAll();
}
Consider a Spring Data Jpa Repository:
public interface UserRepository extends JpaRepository<User, Long> {
User findOneByDeletedIsFalseAndActivationKey(String activationKey);
List<User> findAllByDeletedIsFalseAndActivatedIsFalseAndCreatedDateBefore(DateTime dateTime);
User findOneByDeletedIsFalseAndLogin(String login);
User findOneByDeletedIsFalseAndEmail(String email);
}
Notice each method has "DeletedIsFalse" in it. Is there a simple way to make method names shorter? Like i.e.:
#FullMethodName("findOneByDeletedIsFalseAndEmail")
User findOneByEmail(String email);
Use default Java 8 feature for wrapping, just like so:
interface UserInterface extends JpaRepository<User, Long> {
// use findOneByEmail instead
User findOneByDeletedIsFalseAndEmail(String email);
default User findOneByEmail(String email) {
return findOneByDeletedIsFalseAndEmail(email);
}
}
See an example.
With Kotlin, you can use extension functions, for example:
interface UserRepository : JpaRepository<User, Long> {
// use findOneByEmail instead
fun findOneByDeletedIsFalseAndEmail(email: String): User
}
fun UserRepository.findOneByEmail(email: String) =
findOneByDeletedIsFalseAndEmail(email)
Now you can use Java 8 default interface methods as #Maksim Kostromin described. But there is no such a feature in Spring.
-- Old answer
There is no such a way. You can specify any name for a method and add an annotation #Query with parameter value which holds desired query to database like this:
#Query(value="select u from User u where u.deleted=false and u.email=:email")
User findOneByEmail(#Param("email")String email);
or, with native sql query:
#Query(value="SELECT * FROM users WHERE deleted=false AND email=?1", nativeQuery=true)
User findOneByEmail(String email);
You can also use names that follow the naming convention for queries since #Query annotation will take precedence over query from method name.
#Query docs
Upd:
from Spring docs:
Although getting a query derived from the method name is quite convenient, one might face the situation in which ... the method name would get unnecessarily ugly. So you can either use JPA named queries through a naming convention ... or rather annotate your query method with #Query.
Is there a way to get EF CTP5 to create an index when it creates a schema?
Update: See here for how EF 6.1 handles this (as pointed out by juFo below).
You can take advantage of the new CTP5’s ExecuteSqlCommand method on Database class which allows raw SQL commands to be executed against the database.
The best place to invoke SqlCommand method for this purpose is inside a Seed method that has been overridden in a custom Initializer class. For example:
protected override void Seed(EntityMappingContext context)
{
context.Database.ExecuteSqlCommand("CREATE INDEX IX_NAME ON ...");
}
As some mentioned in the comments to Mortezas answer there is a CreateIndex/DropIndex method if you use migrations.
But if you are in "debug"/development mode and is changing the schema all the time and are recreating the database every time you can use the example mentioned in Morteza answer.
To make it a little easier, I have written a very simple extension method to make it strongly typed, as inspiration that I want to share with anyone who reads this question and maybe would like this approach aswell. Just change it to fit your needs and way of naming indexes.
You use it like this: context.Database.CreateUniqueIndex<User>(x => x.Name);
.
public static void CreateUniqueIndex<TModel>(this Database database, Expression<Func<TModel, object>> expression)
{
if (database == null)
throw new ArgumentNullException("database");
// Assumes singular table name matching the name of the Model type
var tableName = typeof(TModel).Name;
var columnName = GetLambdaExpressionName(expression.Body);
var indexName = string.Format("IX_{0}_{1}", tableName, columnName);
var createIndexSql = string.Format("CREATE UNIQUE INDEX {0} ON {1} ({2})", indexName, tableName, columnName);
database.ExecuteSqlCommand(createIndexSql);
}
public static string GetLambdaExpressionName(Expression expression)
{
MemberExpression memberExp = expression as MemberExpression;
if (memberExp == null)
{
// Check if it is an UnaryExpression and unwrap it
var unaryExp = expression as UnaryExpression;
if (unaryExp != null)
memberExp = unaryExp.Operand as MemberExpression;
}
if (memberExp == null)
throw new ArgumentException("Cannot get name from expression", "expression");
return memberExp.Member.Name;
}
Update: From version 6.1 and onwards there is an [Index] attribute available.
For more info, see http://msdn.microsoft.com/en-US/data/jj591583#Index
This feature should be available in the near-future via data annotations and the Fluent API. Microsoft have added it into their public backlog:
http://entityframework.codeplex.com/workitem/list/basic?keywords=DevDiv [Id=87553]
Until then, you'll need to use a seed method on a custom Initializer class to execute the SQL to create the unique index, and if you're using code-first migrations, create a new migration for adding the unique index, and use the CreateIndex and DropIndex methods in your Up and Down methods for the migration to create and drop the index.
Check my answer here Entity Framework Code First Fluent Api: Adding Indexes to columns this allows you to define multi column indexes by using attributes on properties.