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.
Related
Assuming I've the following endpoints in spring boot
GET /todo
DELETE /todo/{id}
How can ensure that only entries for the userid are returned and that the user can only update his own todos?
I've a populated Authentication object.
Is there any build in way I can use? Or just make sure to always call findXyzByIdAndUserId where userid is always retrieved from the Principal?
I'm a bit worried about the possibility to forget the check and displaying entries from other users.
My approach to this would be a 3 way implementation: (using jpa & hibernate)
a user request context
a mapped superclass to get your context
a statement inspector to inject your userid
For example:
public final class UserRequestContext {
public static String getUserId() {
// code to retrieve your userid and throw when there is none!
if (userId == null) throw new IllegalStateException("userid null");
return userId;
}
}
#MappedSuperclass
public class UserResolver {
public static final String USER_RESOLVER = "USER_RESOLVER";
#Access(AccessType.PROPERTY)
public String getUserId() {
return UserRequestContext.getUserId();
}
}
#Component
public class UserInspector implements StatementInspector {
#Override
public String inspect(String statement) {
if (statement.contains(UserResolver.USER_RESOLVER)) {
statement = statement.replace(UserResolver.USER_RESOLVER, "userId = '" + UserRequestContext.getUserId() + "'" );
}
return sql;
}
#Bean
public HibernatePropertyCustomizer hibernatePropertyCustomizer() {
return hibernateProperies -> hibernateProperties.put("hibernate.session_factory.statement_inspector",
UserInspector.class.getName());
}
}
So your Entity looks like this:
#Entity
...
#Where(clause = UserResolver.USER_RESOLVER)
public class Todo extends UserResolver {
...
}
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);
}
}
I have a class that represents a user date of birth in two separated fields
public class User {
private int yearOfBirth;
private int monthOfBirth;
}
Is it possible to make a projection that exports the user age? I know we can concatenate fields using #Value.
The easiest way to resolve the problem (if you can add code to the domain class) is to add a method in the user class like the one below:
#JsonIgnore
public int getAge() {
return Period.between(
LocalDate.of(dobYear, dobMonth, 1),
LocalDate.now()
).getYears();
}
You can add the #JsonIgnore to block spring from exporting an "age" field when your entity is serialized. After adding that method you can create projection like the one below:
#Projection(name = "userAge ", types = {User.class})
public interface UserAge {
#Value("#{target.getAge()}")
Integer getAge();
}
Something like this, for example:
public class UserAgeDto {
private int yearOfBirth;
private int monthOfBirth;
public UserAgeDto(int yearOfBirth, int monthOfBirth) {
// constructor implementation...
}
public int getAge() {
// age calculation...
}
}
public interface UserRepo extends JpaRepository<User, Long> {
#Query("select new com.example.myapp.dto.UserAgeDto(u.yearOfBirth, u.monthOfBirth) from User u where u = ?")
UserAgeDto getUserAgeDto(User user);
}
Some info
I am using the hibernate validator group sequence and want to execute the groups in a sequence based on business rules. But the input to the groupSequenceProvider for its getValidationGroups is always null, and hence custom sequence never gets added.
My request object:
#GroupSequenceProvider(BeanSequenceProvider.class)
public class MyBean {
#NotEmpty
private String name;
#NotNull
private MyType type;
#NotEmpty(groups = Special.class)
private String lastName;
// Getters and setters
}
Enum type:
public enum MyType {
FIRST, SECOND
}
My custom sequence provider:
public class BeanSequenceProvider implements DefaultGroupSequenceProvider<MyBean> {
#Override
public List<Class<?>> getValidationGroups(MyBean object) {
final List<Class<?>> classes = new ArrayList<>();
classes.add(MyBean.class);
if (object != null && object.getType() == MyType.SECOND) {
classes.add(Special.class);
}
return classes;
}
}
Group annotation:
public interface Special {
}
When I execute the above code, I get the input MyBean object as null and cannot add the custom sequence. What am I missing? I am using hibernate-validator version as 5.4.1.Final
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);
}