Why is just this one field of a Hibernate controlled object/entity not being created or updated?
I have an object "Chemical" that is updated according to entries in an HTML page. The data is returned to the Java code correctly, yet this one field is not updating or being created in the database. The chemical entity is defined as follows:
@Entity
@NamedQuery(name="Chemical.findAll", query="SELECT c FROM Chemical c")
public class Chemical implements Serializable {
@Id
@Column(name="chemical_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int chemicalId;
...
private string formula; <--- THE FIELD THAT WILL NOT UPDATE/BE CREATED
....
public void setFormula(String formula) {
this.formula = formula;
}
public String getFormula() {
return this.formula;
}
.....
The Java code that actually saves/updates the database:
public void saveOrUpdate(final T data) throws CPDPersistenceException {
final EntityManager em = getEntityManager();
try {
final EntityTransaction transaction = em.getTransaction();
transaction.begin();
em.merge(data);
transaction.commit();
} catch (final PersistenceException e) {
throw new PersistenceException(e);
}
}
After the code is executed, other fields have changed in the database if changes have been made. However, "formula" is not changed and remains a NULL field. Can anybody see why?
java hibernate jpa
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
add a comment |
I have an object "Chemical" that is updated according to entries in an HTML page. The data is returned to the Java code correctly, yet this one field is not updating or being created in the database. The chemical entity is defined as follows:
@Entity
@NamedQuery(name="Chemical.findAll", query="SELECT c FROM Chemical c")
public class Chemical implements Serializable {
@Id
@Column(name="chemical_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int chemicalId;
...
private string formula; <--- THE FIELD THAT WILL NOT UPDATE/BE CREATED
....
public void setFormula(String formula) {
this.formula = formula;
}
public String getFormula() {
return this.formula;
}
.....
The Java code that actually saves/updates the database:
public void saveOrUpdate(final T data) throws CPDPersistenceException {
final EntityManager em = getEntityManager();
try {
final EntityTransaction transaction = em.getTransaction();
transaction.begin();
em.merge(data);
transaction.commit();
} catch (final PersistenceException e) {
throw new PersistenceException(e);
}
}
After the code is executed, other fields have changed in the database if changes have been made. However, "formula" is not changed and remains a NULL field. Can anybody see why?
java hibernate jpa
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
You need to add@Column(name="formula")above the variable.
– Jake Stewart
Nov 20 '18 at 20:03
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
Nope, you don't need to define the@Columnon a field ... JPA defines defaults (read the JPA spec if not believing me).@Basicis the most that could be needed but most JPA providers don't require that since String is a supported type
– Billy Frost
Nov 21 '18 at 8:45
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59
add a comment |
I have an object "Chemical" that is updated according to entries in an HTML page. The data is returned to the Java code correctly, yet this one field is not updating or being created in the database. The chemical entity is defined as follows:
@Entity
@NamedQuery(name="Chemical.findAll", query="SELECT c FROM Chemical c")
public class Chemical implements Serializable {
@Id
@Column(name="chemical_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int chemicalId;
...
private string formula; <--- THE FIELD THAT WILL NOT UPDATE/BE CREATED
....
public void setFormula(String formula) {
this.formula = formula;
}
public String getFormula() {
return this.formula;
}
.....
The Java code that actually saves/updates the database:
public void saveOrUpdate(final T data) throws CPDPersistenceException {
final EntityManager em = getEntityManager();
try {
final EntityTransaction transaction = em.getTransaction();
transaction.begin();
em.merge(data);
transaction.commit();
} catch (final PersistenceException e) {
throw new PersistenceException(e);
}
}
After the code is executed, other fields have changed in the database if changes have been made. However, "formula" is not changed and remains a NULL field. Can anybody see why?
java hibernate jpa
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
I have an object "Chemical" that is updated according to entries in an HTML page. The data is returned to the Java code correctly, yet this one field is not updating or being created in the database. The chemical entity is defined as follows:
@Entity
@NamedQuery(name="Chemical.findAll", query="SELECT c FROM Chemical c")
public class Chemical implements Serializable {
@Id
@Column(name="chemical_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int chemicalId;
...
private string formula; <--- THE FIELD THAT WILL NOT UPDATE/BE CREATED
....
public void setFormula(String formula) {
this.formula = formula;
}
public String getFormula() {
return this.formula;
}
.....
The Java code that actually saves/updates the database:
public void saveOrUpdate(final T data) throws CPDPersistenceException {
final EntityManager em = getEntityManager();
try {
final EntityTransaction transaction = em.getTransaction();
transaction.begin();
em.merge(data);
transaction.commit();
} catch (final PersistenceException e) {
throw new PersistenceException(e);
}
}
After the code is executed, other fields have changed in the database if changes have been made. However, "formula" is not changed and remains a NULL field. Can anybody see why?
java hibernate jpa
java hibernate jpa
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
asked Nov 20 '18 at 19:50
calvinnmecalvinnme
74
74
You need to add@Column(name="formula")above the variable.
– Jake Stewart
Nov 20 '18 at 20:03
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
Nope, you don't need to define the@Columnon a field ... JPA defines defaults (read the JPA spec if not believing me).@Basicis the most that could be needed but most JPA providers don't require that since String is a supported type
– Billy Frost
Nov 21 '18 at 8:45
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59
add a comment |
You need to add@Column(name="formula")above the variable.
– Jake Stewart
Nov 20 '18 at 20:03
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
Nope, you don't need to define the@Columnon a field ... JPA defines defaults (read the JPA spec if not believing me).@Basicis the most that could be needed but most JPA providers don't require that since String is a supported type
– Billy Frost
Nov 21 '18 at 8:45
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59
You need to add
@Column(name="formula") above the variable.– Jake Stewart
Nov 20 '18 at 20:03
You need to add
@Column(name="formula") above the variable.– Jake Stewart
Nov 20 '18 at 20:03
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
Nope, you don't need to define the
@Column on a field ... JPA defines defaults (read the JPA spec if not believing me). @Basic is the most that could be needed but most JPA providers don't require that since String is a supported type– Billy Frost
Nov 21 '18 at 8:45
Nope, you don't need to define the
@Column on a field ... JPA defines defaults (read the JPA spec if not believing me). @Basic is the most that could be needed but most JPA providers don't require that since String is a supported type– Billy Frost
Nov 21 '18 at 8:45
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59
add a comment |
1 Answer
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oldest
votes
There is I think a difference between UPDATE and CREATE.
Try to replace em.merge(data) with em.persist(data)
Does it save the content the first time you create a new row with persist ?
What is the length of your string definition in the database ?
Check the database logs whenever the HQL is run from your code, do the log complain at one point during the persist ?
It might be necessary to first test if the row exists before you can run a merge.
answered Nov 20 '18 at 21:04
ApolloApollo
362
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is I think a difference between UPDATE and CREATE.
Try to replace em.merge(data) with em.persist(data)
Does it save the content the first time you create a new row with persist ?
What is the length of your string definition in the database ?
Check the database logs whenever the HQL is run from your code, do the log complain at one point during the persist ?
It might be necessary to first test if the row exists before you can run a merge.
answered Nov 20 '18 at 21:04
ApolloApollo
362
add a comment |
There is I think a difference between UPDATE and CREATE.
Try to replace em.merge(data) with em.persist(data)
Does it save the content the first time you create a new row with persist ?
What is the length of your string definition in the database ?
Check the database logs whenever the HQL is run from your code, do the log complain at one point during the persist ?
It might be necessary to first test if the row exists before you can run a merge.
answered Nov 20 '18 at 21:04
ApolloApollo
362
add a comment |
There is I think a difference between UPDATE and CREATE.
Try to replace em.merge(data) with em.persist(data)
Does it save the content the first time you create a new row with persist ?
What is the length of your string definition in the database ?
Check the database logs whenever the HQL is run from your code, do the log complain at one point during the persist ?
It might be necessary to first test if the row exists before you can run a merge.
answered Nov 20 '18 at 21:04
ApolloApollo
362
There is I think a difference between UPDATE and CREATE.
Try to replace em.merge(data) with em.persist(data)
Does it save the content the first time you create a new row with persist ?
What is the length of your string definition in the database ?
Check the database logs whenever the HQL is run from your code, do the log complain at one point during the persist ?
It might be necessary to first test if the row exists before you can run a merge.
answered Nov 20 '18 at 21:04
ApolloApollo
362
answered Nov 20 '18 at 21:04
ApolloApollo
362
answered Nov 20 '18 at 21:04
ApolloApollo
362
answered Nov 20 '18 at 21:04
ApolloApollo
362
362
add a comment |
add a comment |
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You need to add
@Column(name="formula")above the variable.– Jake Stewart
Nov 20 '18 at 20:03
I'll try that, but there are a bunch of other fields that are being populated correctly that do not have the @Column(name = WHATEVER) line above them.
– calvinnme
Nov 20 '18 at 20:31
Nope, you don't need to define the
@Columnon a field ... JPA defines defaults (read the JPA spec if not believing me).@Basicis the most that could be needed but most JPA providers don't require that since String is a supported type– Billy Frost
Nov 21 '18 at 8:45
Also your post doesnt say whether the object is already persistent, whether you created the schema before adding that field, what the schema is, what SQL is invoked when you invoke a JPA operation. All of this is basic info for debugging
– Billy Frost
Nov 21 '18 at 8:59