I have a UserAssignmentRole class like this :
#Data
#Entity
public class UserAssignmentRole {
...
#Enumerated(EnumType.STRING)
public Role role;
}
And the Role is enum, it looks like this:
public enum Role{
admin,
member,
pending
}
Now when in my repository I try to query to select all with role of admin, it gives me error:
#Query("select uar from UserAssignmentRole uar where uar.role=Role.admin")
public List<UserAssignmentRole> listAdmin(Long userID, Long assignmentID);
How this can be solved?
Error : org.hibernate.hql.internal.ast.QuerySyntaxException: Invalid path: 'Role.admin'
Full error : https://pastebin.com/tk9r3wDg
It is a strange but intended behaviour of Hibernate since 5.2.x
An enum value is a constant and you're using a non-conventional naming (lowercase)
Take a look at this issue and Vlad Mihalcea's long explanation of the performance penalty.
If you’re using non-conventional Java constants, then you’ll have to set the hibernate.query.conventional_java_constants configuration property to false. This way, Hibernate will fall back to the previous behavior, treating any expression as a possible candidate for a Java constant.
You can try not to write this sql by yourself but with repository create code like this:
#Repository
public interface UserAssignmentRolelRepository extends JpaRepository<UserModel, Long>{
public List<UserAssignmentRole> findByRole(Role role);
}
And then:
#Autowired
UserAssignmentRolelRepository repository ;
public void someMethod(){
List<UserAssignmentRole> userAssignmentRoles = repository.findByRole(Role.admin);
}
UPDATE 1
As it was point out in this answer: non-conventional naming. You can change labels in your enum to uppercase.
public enum Role{
Admin,
Member,
Pending
}
and then:
#Query("select uar from UserAssignmentRole uar where uar.role=com.example.package.Role.Admin")
public List<UserAssignmentRole> listAdmin(Long userID, Long assignmentID);
UPDATE 2
But if you really want to have lowercase in DB.
It requires more code to change. Enum change to:
public enum Role{
Admin("admin"),
Member("member"),
Pending("pending");
private String name;
Role(String name) {
this.name = name;
}
public String getName() { return name; }
public static Role parse(String id) {
Role role = null; // Default
for (Role item : Role.values()) {
if (item.name.equals(id)) {
role = item;
break;
}
}
return role;
}
}
In UserAssignmentRole
// #Enumerated(EnumType.STRING)
#Convert(converter = RoleConverter.class)
private Role role;
And additional class:
import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
#Converter(autoApply = true)
public class RoleConverter implements AttributeConverter<Role, String> {
#Override
public String convertToDatabaseColumn(Role role) {
return role.getName();
}
#Override
public Role convertToEntityAttribute(String dbData) {
return Role.parse(dbData);
}
}
Related
is it posible to generate a custom "presence checking" method name, being a method of the property itself rather the owning object?
I know I can use hasProperty() methods to check for presence of a value...
https://mapstruct.org/documentation/stable/reference/html/#source-presence-check
but with Optional or JsonNullable (from OpenApi nonullable) that checking method is on the property itself, not on the owning object... :-(
I can map JsonNullable or Optional easyly 'using' or extending a simple custom Mapper
#Mapper
public class JsonNullableMapper {
public <T> T fromJsonNullable(final JsonNullable<T> jsonNullable) {
return jsonNullable.orElse(null);
}
public <T> JsonNullable<T> asJsonNullable(final T nullable) {
return nullable != null ? JsonNullable.of(nullable) : JsonNullable.undefined();
}
}
what I would like to achieve is something like this as "presence check":
if(source.getProperty().isPresent()) {
target.set(customMapper.map(source.getProperty()));
}
Any one found a solution for this?
Thanks and regards
I have managed to implement custom lombok extension which generates "presence checknig" methods.
Here is an example project. In short I added #PresenceChecker annotation and implemented Lombok Javac Annotation handler.
It's possible to use it together with other Lombok annotations:
#Getter
#Setter
public class User {
private String name;
}
#Getter
#Setter
#PresenceChecker
public class UserUpdateDto {
private String name;
}
//MapStruct Mapper interface declaration
#Mapper
public interface UserMapper {
void updateUser(UserUpdateDto dto, #MappingTarget User user);
}
Generated code:
public class User {
private String name;
public String getName() {
return this.name;
}
public void setName(String name) {
this.name = name;
}
}
public class UserUpdateDto {
private boolean hasName;
private String name;
public String getName() {
return this.name;
}
public void setName(String name) {
this.name = name;
this.hasName = true;
}
public boolean hasName() {
return this.hasName;
}
}
//MapStruct Mapper implementation
public class UserMapperImpl implements UserMapper {
#Override
public void updateUser(UserUpdateDto dto, User user) {
if ( dto == null ) {
return;
}
if ( dto.hasName() ) {
user.setName( dto.getName() );
}
}
}
The answer is unfortunately a straight no.
It is not possible in the current version of MapStruct (1.3.1final) and its not on the shortlist for 1.4.0. You could open up an issue on the git repo of MapStruct as feature request.
I have a Role enum, like this:
public enum Role{
admin('a'),
member('m'),
pending('p');
char role;
Role(char a) {
this.role = a;
}
public char getRole() {
return role;
}
public static Role getByRole(char role) {
return Arrays.stream(Role.values())
.filter(Role -> Role.getRole() == role)
.findFirst()
.orElse(Role.pending);
}
}
To support conversion, I have created a class called RoleConverter:
#Converter
public class RoleConverter implements AttributeConverter<Role, Character> {
#Override
public Character convertToDatabaseColumn(Role Role) {
return Role.getRole();
}
#Override
public Role convertToEntityAttribute(Character dbData) {
System.out.println(dbData);
return Role.getByRole(dbData);
}
}
And in my Target object I have added proper annotations:
#Convert(converter = RoleConverter.class)
#Enumerated(EnumType.STRING)
public Role role;
Still it gives me error - nested exception is org.springframework.dao.InvalidDataAccessApiUsageException: No enum constant com.mua.cse616.model.Role.2;
Using spring with h2 and jpa
Seems like you have a row in your DB which has in the column the value 2 which obviously is not present in the enum. Maybe you started out without the #Enumerated annotation thus JPA used the ordinal as the column value.
Your database contains an entry with role = 2.
Make sure that the entries in the database have the same values as in your Enum.
I user a CRUDRepository in my spring data redis project to persist a redis hash in my redis cluster. i have rest api written to persist and get thte values of the data. this works fine.
however my entity annotated with RedisHash is being saved as a set / and i am not able to look inside the value using redis cli.
how do i look inside a set data type(without popping) in redis cli
i looked at redis commands page https://redis.io/commands#set
i only get operations which can pop value . i neeed to simply peek
EDIT:
to make things clearer, i am using spring crudrepo to save the user entity into redis data store. the user entity gets saved as a set data type.
when i query back the user details, i can see entire details of the user
{
userName: "somak",
userSurName: "dattta",
age: 23,
zipCode: "ah56h"
}
i essentially want to do the same using redis cli... but all i get is
127.0.0.1:6379> smembers user
1) "somak"
how do i look inside the somak object.
#RestController
#RequestMapping("/immem/core/user")
public class UserController {
#Autowired
private UserRepository userRepository;
#RequestMapping(path = "/save", method = RequestMethod.GET, produces = "application/json")
#ResponseStatus(HttpStatus.OK)
public void saveUserDetails() {
User user = new User();
user.setAge(23);
user.setUserName("somak");
user.setUserSurName("dattta");
user.setZipCode("ah56h");
userRepository.save(user);
}
#RequestMapping(path="/get/{username}", method = RequestMethod.GET, produces = "application/json")
public User getUserDetails(#PathVariable("username") String userName) {
return userRepository.findById(userName).get();
}
}
#Repository
public interface UserRepository extends CrudRepository<User, String>{
}
#RedisHash("user")
public class User {
private #Id String userName;
private #Indexed String userSurName;
private #Indexed int age;
private String zipCode;
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getUserSurName() {
return userSurName;
}
public void setUserSurName(String userSurName) {
this.userSurName = userSurName;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getZipCode() {
return zipCode;
}
public void setZipCode(String zipCode) {
this.zipCode = zipCode;
}
}
I don't understant your descr with your problem, but I understand your title.
In redis set, the member is always string type.
I hope you can offer more info about UserRepository.save:
User user = new User();
user.setAge(23);
user.setUserName("somak");
user.setUserSurName("dattta");
user.setZipCode("ah56h");
userRepository.save(user);
And you can check your redis data and check data type when rest api invoked.
I have the following MongoDB Repository
public interface TeamRepository extends MongoRepository<Team, TeamId> {
....
}
And the following classes:
public abstract class DbId implements Serializable {
#Id
private final String id;
public DbId(final String id) { this.id = id;}
public String getId() { return id;}
}
public class TeamId extends DbId {
public TeamId(final String id) {
super(id)
}
}
As you can see, I have like a custom id for the repository (I have MongoRepository instead of something like MongoRepository). But, when I am trying to save a Team object, I get an error saying that MongoDB does not know how to generate DBId. Any clue?
MongoDb (or any database) would not know how to generate a string ID without you informing it what the value of the string is.
The default #Id is a string representation of ObjectId, which can be auto-generated by MongoDB. If you are changing the type of string ObjectId to a class, then at least the class needs to define:
** Conversion to string (serialisable), for example:
#Override
public String toString() {
return String.format(
"TeamID[uniqueString=%s]",
myUniqueString);
}
** How to generate the Id.
You can define a method in your TeamRepository i.e. save() to specify how your string can be generated. Alternatively you can check out
https://www.mkyong.com/mongodb/spring-data-mongodb-auto-sequence-id-example/
Where the example specify getNextSequenceId() to generate NumberLong custom id. Hopefully that guides you to your answer.
I have an error in spring JPA
org.springframework.data.mapping.PropertyReferenceException: No property CompanyId found for type CompanyUserDetail!
#Embeddable
public class CompanyUserKey implements Serializable {
public CompanyUserKey() {
}
#Column(name = "company_id")
private UUID companyId;
#Column(name = "user_name")
private String userName;
public UUID getCompanyId() {
return companyId;
}
public void setCompanyId(UUID companyId) {
this.companyId = companyId;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
}
#Entity
#Table(name = "company_user_detail")
public class CompanyUserDetail {
#EmbeddedId
CompanyUserKey companyUserkey;
public CompanyUserKey getCompanyUserkey() {
return companyUserkey;
}
public void setCompanyUserkey(CompanyUserKey companyUserkey) {
this.companyUserkey = companyUserkey;
}
}
I am trying to access below method Service layer
#Component
public interface CompanyUserRepository extends JpaRepository<CompanyUserDetail, CompanyUserKey> {
public List<CompanyUserDetail> findByCompanyId(UUID companyId);
}
How can I achieve this ?
Thanks
Since in java model your CompanyUserKey is a property in the CompanyUserDetail class, I believe you should use full path (companyUserkey.companyId) to reach companyId:
public List<CompanyUserDetail> findByCompanyUserkeyCompanyId(UUID companyId);
Also note that you have a naming inconsistency: field in CompanyUserDetail is named companyUserkey instead of companyUserKey.
Assuming you are not using spring-data-jpa's auto generated implementations, your method contents might look something like the following:
FROM CompanyUserDetail c WHERE c.companyUserKey.companyId = :companyId
Now simply provide that query to the EntityManager
entityManager.createQuery( queryString, CompanyUserDetail.class )
.setParameter( "companyId", companyId )
.getResultList();
The key points are:
Query uses a named bind parameter called :companyId (not the leading :).
Parameter values are bound in a secondary step using setParameter method variants.
createQuery uses a second argument to influence type safety so that the return value from getResultList is a List<CompanyUserDetail> just like you requested.
Looking at spring-data-jpa's implementation however, I suspect it could look like this:
public interface CustomerUserRepository
extends JpaRepository<CompanyUserDetail, CompanyUserKey> {
#Query("select c FROM CompanyUserDetail c WHERE c.companyUserKey.companyId = :companyId")
List<CompanyUserDetail> findByCompanyId(#Param("companyId") UUID companyId);
}