Saving integer array using Spring Data JPA
I am using spring data jpa with hibernate and postgres. I am trying to save an integer array in a column. I am using vlad mihaceas library for persisting the array into postgresql. The entity is as follows:-
@Type(type = "int-array")
@Column(name = "location", columnDefinition = "integer")
private Integer locations;
The corresponding location entity is
@Entity
@Table(name = "location_master")
public class Location implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column(name = "location_name")
private String locationName;
}
The location ids are being saved in the db. But I am not able to display the in thymeleaf.
<tr>
<td>Selected Locations</td>
<td>[[${office.locations[0].locationName}]]</td>
</tr>
The following error has come up:-
org.springframework.expression.spel.SpelEvaluationException: EL1008E:
Property or field 'locationName' cannot be found on object of type
'java.lang.Integer' - maybe not public or not valid?
java spring postgresql hibernate spring-data-jpa
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
|
show 1 more comment
I am using spring data jpa with hibernate and postgres. I am trying to save an integer array in a column. I am using vlad mihaceas library for persisting the array into postgresql. The entity is as follows:-
@Type(type = "int-array")
@Column(name = "location", columnDefinition = "integer")
private Integer locations;
The corresponding location entity is
@Entity
@Table(name = "location_master")
public class Location implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column(name = "location_name")
private String locationName;
}
The location ids are being saved in the db. But I am not able to display the in thymeleaf.
<tr>
<td>Selected Locations</td>
<td>[[${office.locations[0].locationName}]]</td>
</tr>
The following error has come up:-
org.springframework.expression.spel.SpelEvaluationException: EL1008E:
Property or field 'locationName' cannot be found on object of type
'java.lang.Integer' - maybe not public or not valid?
java spring postgresql hibernate spring-data-jpa
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
1
Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
Try to use@ElementCollectionand@CollectionTableannotations
– Nick Savenia
Nov 15 '18 at 9:55
|
show 1 more comment
I am using spring data jpa with hibernate and postgres. I am trying to save an integer array in a column. I am using vlad mihaceas library for persisting the array into postgresql. The entity is as follows:-
@Type(type = "int-array")
@Column(name = "location", columnDefinition = "integer")
private Integer locations;
The corresponding location entity is
@Entity
@Table(name = "location_master")
public class Location implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column(name = "location_name")
private String locationName;
}
The location ids are being saved in the db. But I am not able to display the in thymeleaf.
<tr>
<td>Selected Locations</td>
<td>[[${office.locations[0].locationName}]]</td>
</tr>
The following error has come up:-
org.springframework.expression.spel.SpelEvaluationException: EL1008E:
Property or field 'locationName' cannot be found on object of type
'java.lang.Integer' - maybe not public or not valid?
java spring postgresql hibernate spring-data-jpa
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
I am using spring data jpa with hibernate and postgres. I am trying to save an integer array in a column. I am using vlad mihaceas library for persisting the array into postgresql. The entity is as follows:-
@Type(type = "int-array")
@Column(name = "location", columnDefinition = "integer")
private Integer locations;
The corresponding location entity is
@Entity
@Table(name = "location_master")
public class Location implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column(name = "location_name")
private String locationName;
}
The location ids are being saved in the db. But I am not able to display the in thymeleaf.
<tr>
<td>Selected Locations</td>
<td>[[${office.locations[0].locationName}]]</td>
</tr>
The following error has come up:-
org.springframework.expression.spel.SpelEvaluationException: EL1008E:
Property or field 'locationName' cannot be found on object of type
'java.lang.Integer' - maybe not public or not valid?
java spring postgresql hibernate spring-data-jpa
java spring postgresql hibernate spring-data-jpa
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
edited Nov 24 '18 at 6:57
Vlad Mihalcea
56.8k13153452
56.8k13153452
asked Nov 15 '18 at 7:44
JohnJohn
1541312
asked Nov 15 '18 at 7:44
JohnJohn
1541312
asked Nov 15 '18 at 7:44
JohnJohn
1541312
1541312
1
Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
Try to use@ElementCollectionand@CollectionTableannotations
– Nick Savenia
Nov 15 '18 at 9:55
|
show 1 more comment
1
Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
Try to use@ElementCollectionand@CollectionTableannotations
– Nick Savenia
Nov 15 '18 at 9:55
1
1
Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
Try to use
@ElementCollection and @CollectionTable annotations– Nick Savenia
Nov 15 '18 at 9:55
Try to use
@ElementCollection and @CollectionTable annotations– Nick Savenia
Nov 15 '18 at 9:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
org.springframework.expression.spel.SpelEvaluationException: EL1008E: Property or field 'locationName' cannot be found on object of type 'java.lang.Integer' - maybe not public or not valid?
The error has nothing to do with the use of the int-array of the hibernate-types project.
The error message is about the Spring Expression Language you used for setting the locationName
which is a String property.
[[${office.locations[0].locationName}]]
The locations property is an Integer, but you treated it as a Location array.
What you want is a @OneToMany List<Location> instead:
@OneToMany(mappedBy="office", cascade = CascadeType.ALL, orphanRemoval = true)
private List<Location> locations = new ArrayList<>();
And a @ManyToOne association in Location:
@ManyToOne(fetch = FetchType.LAZY)
private Office office;
Check out this article for more details about the best way to use a @OneToMany association.
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
|
show 1 more comment
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org.springframework.expression.spel.SpelEvaluationException: EL1008E: Property or field 'locationName' cannot be found on object of type 'java.lang.Integer' - maybe not public or not valid?
The error has nothing to do with the use of the int-array of the hibernate-types project.
The error message is about the Spring Expression Language you used for setting the locationName
which is a String property.
[[${office.locations[0].locationName}]]
The locations property is an Integer, but you treated it as a Location array.
What you want is a @OneToMany List<Location> instead:
@OneToMany(mappedBy="office", cascade = CascadeType.ALL, orphanRemoval = true)
private List<Location> locations = new ArrayList<>();
And a @ManyToOne association in Location:
@ManyToOne(fetch = FetchType.LAZY)
private Office office;
Check out this article for more details about the best way to use a @OneToMany association.
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
|
show 1 more comment
org.springframework.expression.spel.SpelEvaluationException: EL1008E: Property or field 'locationName' cannot be found on object of type 'java.lang.Integer' - maybe not public or not valid?
The error has nothing to do with the use of the int-array of the hibernate-types project.
The error message is about the Spring Expression Language you used for setting the locationName
which is a String property.
[[${office.locations[0].locationName}]]
The locations property is an Integer, but you treated it as a Location array.
What you want is a @OneToMany List<Location> instead:
@OneToMany(mappedBy="office", cascade = CascadeType.ALL, orphanRemoval = true)
private List<Location> locations = new ArrayList<>();
And a @ManyToOne association in Location:
@ManyToOne(fetch = FetchType.LAZY)
private Office office;
Check out this article for more details about the best way to use a @OneToMany association.
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
|
show 1 more comment
org.springframework.expression.spel.SpelEvaluationException: EL1008E: Property or field 'locationName' cannot be found on object of type 'java.lang.Integer' - maybe not public or not valid?
The error has nothing to do with the use of the int-array of the hibernate-types project.
The error message is about the Spring Expression Language you used for setting the locationName
which is a String property.
[[${office.locations[0].locationName}]]
The locations property is an Integer, but you treated it as a Location array.
What you want is a @OneToMany List<Location> instead:
@OneToMany(mappedBy="office", cascade = CascadeType.ALL, orphanRemoval = true)
private List<Location> locations = new ArrayList<>();
And a @ManyToOne association in Location:
@ManyToOne(fetch = FetchType.LAZY)
private Office office;
Check out this article for more details about the best way to use a @OneToMany association.
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
org.springframework.expression.spel.SpelEvaluationException: EL1008E: Property or field 'locationName' cannot be found on object of type 'java.lang.Integer' - maybe not public or not valid?
The error has nothing to do with the use of the int-array of the hibernate-types project.
The error message is about the Spring Expression Language you used for setting the locationName
which is a String property.
[[${office.locations[0].locationName}]]
The locations property is an Integer, but you treated it as a Location array.
What you want is a @OneToMany List<Location> instead:
@OneToMany(mappedBy="office", cascade = CascadeType.ALL, orphanRemoval = true)
private List<Location> locations = new ArrayList<>();
And a @ManyToOne association in Location:
@ManyToOne(fetch = FetchType.LAZY)
private Office office;
Check out this article for more details about the best way to use a @OneToMany association.
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
answered Nov 24 '18 at 6:57
Vlad MihalceaVlad Mihalcea
56.8k13153452
56.8k13153452
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
|
show 1 more comment
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
I know that it is a SpEL error and nothing to do with hibernate-types. Perhaps i could not explain my problem properly. The OneToMany List<Location mapping is also fine with me. My question is whether the location ids will be stored in the database as an integer array by this association.
– John
Nov 24 '18 at 7:26
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
Of course, they will. Check out the hibernate-types tests in the GitHub repository for a proof.
– Vlad Mihalcea
Nov 24 '18 at 8:09
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I don't need proof. Hibernate types is already working fine for me in jsonb types. I am only confused as how to use @Type type= int-array annotation in these mappings. You also didn't mention it in your answer.
– John
Nov 24 '18 at 9:04
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
I would have mentioned it the exception message would indicate that. Check out this test case to see how to use the int-array mapping. Just compare my test case with yours and see where they differ.
– Vlad Mihalcea
Nov 24 '18 at 11:44
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
Yes, i have seen it. You have an array of integers and in my case, it is a list of objects. I am not able to figure out how to do the int-array mapping for this list
– John
Nov 24 '18 at 11:58
|
show 1 more comment
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Well, the message says it all. Why do you think you can get a locationName out of an Integer? That makes no sense. An Integer is an Integer. Not a Location. If you want a Collection<Location>, then use a Collection<Location>, and make it a OneToMany association between your unknown entity and the Location entity.
– JB Nizet
Nov 15 '18 at 7:49
@JB Nizet I want to save the selected location ids in the database as array. I already tried with List<Integer> which gave the error Use of OneToMany or ManyToMany targeting an unmapped class. As Hibernate maps only primitive types, i was using vlad mihalcea's library
– John
Nov 15 '18 at 8:46
That's not proper database design. Use foreign keys, join columns, join tables, and associations between entities.
– JB Nizet
Nov 15 '18 at 9:36
@JBNizet Database design is not in my control. I have to work with what is given to me.
– John
Nov 15 '18 at 9:51
Try to use
@ElementCollectionand@CollectionTableannotations– Nick Savenia
Nov 15 '18 at 9:55