I am trying to retrieve a USER from my database using the ID in the WHERE clause. However I am receiving an error and my program is failing.
This is the error I'm receiving when I run my program:
ERROR [org.jboss.as.ejb3.invocation] (default task-19)
JBAS014134: EJB Invocation failed on component CustomerServiceBeanImpl
for method public abstract package.name.entity.ICustomer
package.name.bean.CustomerServiceBean.getCustomerById(long):
javax.ejb.EJBException: java.lang.IllegalArgumentException:
Parameter value [19533] did not match expected type [package.name.entity.User (n/a)]
Note: [19533] is a test value I used.
This is the method that is having the error in the CustomerServiceBeanImpl.java:
#Override
public Customer getCustomerById (final long id)
{
return Customer.getById (this.em, id);
}
This is the method that's being called by the component CustomerServiceBeanImpl:
public static Customer getById (final EntityManager em, final long id)
{
for (final Customer c : em.createNamedQuery ("Customer.getById", Customer.class)
.setParameter ("id", id).setMaxResults (1).getResultList ())
{
return c;
}
return null;
}
The name query is this:
#NamedQuery (name = "Customer.getById",
query = "SELECT o FROM gnf.Customer o WHERE o.user = :id")
In the Customer.java class itself the relevant column is this one:
#ManyToOne (fetch = FetchType.LAZY)
#JoinColumn (name = "user_id")
private User user;
Doing a quick check of my ERD the "id" column in my "customer" table has a data type of BIGINT. However I'm not sure if that matters. (PostgreSQL database by the way.)
How can I fix this error?
The WHERE clause in your named query seems to be the problem. The attribute user in your Customer.class is of type User and your query expects it to be a type compatible to long.
...
Parameter value [19533] did not match expected type [package.name.entity.User ...
So if you need more help on this it would be great to see the complete entities User and Customer.
It is happening because in your database the parameter will be a #oneToOne object and you have to call the id inside the Object so you have to give the query as following :
"select user from User user where user.customer.id=:param"
Related
I have two classes, Account and Admin, with many to many mapping.
The Admin class has a collection of Account class and vise versa.
I want to write a query, that given the account id, will return all the account admins.
Here is the relevant fields of the Account class:
#Entity
public class Account {
#Id
public Long id;
#ManyToMany(mappedBy = "account", fetch = FetchType.LAZY, cascade = CascadeType.ALL)
public List<Admin> users = new ArrayList<>();
}
I have tried a regular query for Admin.class with multiselect as each account has a collection of admins, but trying to get a TypedQuery<Admin> out of my CriteriaQuery<Admin> I got an IllegalArgumentException with the message "org.hibernate.hql.internal.ast.QuerySyntaxException: Unable to locate appropriate constructor on class [models.Admin]. Expected arguments are: java.util.Collection [select new models.Admin(generatedAlias0.users) from models.Account as generatedAlias0 where generatedAlias0.id=1L]" (1L here probably since I called this function with 1 as accountId), caused by QuerySyntaxException with the message "Unable to locate appropriate constructor on class [models.Admin]. Expected arguments are: java.util.Collection".
Code:
private static List<Admin> readAccountAdmins(Long accountId) {
CriteriaBuilder cb = JPA.em().getCriteriaBuilder();
CriteriaQuery<Admin> cq = cb.createQuery(Admin.class);
Root<Account> root = cq.from(Account.class);
Predicate idPredicate = cb.equal(root.get(Account_.id), accountId);
cq.multiselect(root.get(Account_.users)).where(idPredicate);
TypedQuery<Admin> typedQuery = JPA.em().createQuery(cq); // exception thrown here
return typedQuery.getResultList();
}
After that I tried running a TypedQuery<List<Admin>>, as I am trying to read a list. This is the first iteration of trying a query of list:
private static List<Admin> readAccountAdmins(Long accountId) {
CriteriaBuilder cb = JPA.em().getCriteriaBuilder();
CriteriaQuery<List<Admin>> cq = cb.createQuery((Class<List<Admin>>)(Class<?>)(Collection.class));
Root<Account> root = cq.from(Account.class);
Predicate idPredicate = cb.equal(root.get(Account_.id), accountId);
cq.select(root.get(Account_.users)).where(idPredicate);
TypedQuery<List<Admin>> typedQuery = JPA.em().createQuery(cq);
return typedQuery.getSingleResult(); // exception thrown here
}
I used getSingleResult as getResultList caused a compilation error, saying the actual return value is List<List<Admin>>> and doesn't match the signature.
This method threw a different exception, a NonUniqueResultException with the message: "result returns more than one elements".
While debugging, I tried to evaluate the expression typedQuery.getResultList() and saw that it actually returns List<Admin> and not List<List<Admin>>, so I got to my final iteration of this function:
private static List<Admin> readAccountAdmins(Long accountId) {
CriteriaBuilder cb = JPA.em().getCriteriaBuilder();
CriteriaQuery<List<Admin>> cq = cb.createQuery((Class<List<Admin>>)(Class<?>)(Collection.class));
Root<Account> root = cq.from(Account.class);
Predicate idPredicate = cb.equal(root.get(Account_.id), accountId);
cq.select(root.get(Account_.users)).where(idPredicate);
TypedQuery<List<Admin>> typedQuery = JPA.em().createQuery(cq);
return (List) typedQuery.getResultList();
}
Now, this function works, but my question is why?
Why did the compiler decide that getResultList returns a different value than the actual return value?
Maybe it makes sense when you take a closer look at your database. A TypeQuery returns entities, so basically rows from tables. List<Admin> is a collection of Entities, so eventhough your Account has a List<Admin> as a field, the Query will still return List<Admin> entities, not List<List<Admin>> as List<Admin> is not an entity.
I hope that makes sense.
I have two tables bo_operator and hist_bo_operator_password. In bo_operator the id column is foreign key to hist_bo_operator_password and I can have many the same operator_id in hist_bo_operator_password and only one id in bo_operator.
My entity:
#Entity
#Table(name="bo_operator")
public class Operator implements Serializable
and that's how I am getting values from hist_bo_operator_password:
#ElementCollection
#CollectionTable(name="hist_bo_operator_password", joinColumns=#JoinColumn(name="id_operator"))
#Column(name="password")
public List<String> oldPasswords = new ArrayList<String>();
but when I'm trying to get only one value by:
#ElementCollection
#CollectionTable(name="hist_bo_operator_password", joinColumns=#JoinColumn(name="id_operator"))
#Column(name="password")
public String oldPassword;
I'm getting error:
org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaAutoConfiguration.class]: Invocation of init method failed; nested exception is org.hibernate.AnnotationException: Illegal attempt to map a non collection as a #OneToMany, #ManyToMany or #CollectionOfElements: local.vlex.operator.model.Operator.oldPassword
and all I want to do is makeing an insert into hist_bo_operator_password by
operator.setOldPassword(oldPassword);. I think the problem is that it doesn't know which password take if there is many values for the same id.
How to achive it?
#Edit
I also tried:
#Table(name="bo_operator")
#SecondaryTable(name = "hist_bo_operator_password",pkJoinColumns=#PrimaryKeyJoinColumn(name="id_operator", referencedColumnName="id"))
I even found ORDER BY so:
#Column(name="password", table="hist_bo_operator_password")
#OrderBy("data_ins")
public String oldPassword;
but seems like there is no #Limit or something like this in JPA and I still have many values to the same id which cause error:
org.hibernate.HibernateException: Duplicate identifier in table for: [local.vlex.operator.model.Operator#1]
As you described at the first sentence you have one-to-many relationship
#ElementCollection
#CollectionTable(name="hist_bo_operator_password", joinColumns=#JoinColumn(name="id_operator"))
#Column(name="password")
#OrderBy("data_ins")
public List<String> oldPasswords = new ArrayList<String>();
Then add necessary getter
Optional<String> getPassword() {
return oldPasswords.size() > 0
? Optional.of(oldPasswords.get(0))
: Optional.empty();
}
And setter
void setPassword(String password) { // or maybe addPassword?
oldPasswords.add(password);
}
Why don't you create entity of hist_bo_operator_password?
Then you could have List of that entities (instead of just String List) and just add another object to List and save entity Operator.
I am trying to return a string from a table based on a conditional ID (subid) from an already populated table. The query should return a list of type ItemDataPoint entity. In a JSF managed bean, the list will the be iterated by a an enhaned for loop. If the word "Include" is found by the loop, the method will create a specific type of chart. In simpler terms, I want to return a string based the ID condition being met. I am getting:
javax.ejb.EJBException
at com.sun.ejb.containers.EJBContainerTransactionManager.processSystemException
(EJBContainerTransactionManager.java:748)
at com.sun.ejb.containers.EJBContainerTransactionManager.
completeNewTx(EJBContainerTransactionManager.java:698)
at com.sun.ejb.containers.EJBContainerTransactionManager.postInvokeTx
(EJBContainerTransactionManager.java:503)
at com.sun.ejb.containers.BaseContainer.postInvokeTx(BaseContainer.java:4475)
at com.sun.ejb.containers.BaseContainer.postInvoke(BaseContainer.java:2009)
at com.sun.ejb.containers.BaseContainer.postInvoke(BaseContainer.java:1979)
Caused by: java.lang.IllegalArgumentException: You have attempted to set
a parameter at position 2 which does not exist in this query string SELECT p FROM
Itemdatapoint p JOIN p.series s WHERE s.master.item.subs.subid = :subid.
at org.eclipse.persistence.internal.jpa.QueryImpl.setParameterInternal(QueryImpl.java:925)
at org.eclipse.persistence.internal.jpa.QueryImpl.setParameterInternal(QueryImpl.java:906)
at org.eclipse.persistence.internal.jpa.EJBQueryImpl.setParameter(EJBQueryImpl.java:469)
at org.eclipse.persistence.internal.jpa.EJBQueryImpl.setParameter(EJBQueryImpl.java:1)
at com.manaar.clientmods.gazprom.design3.data.facade.ItemdatapointFacade.
chartType(ItemdatapointFacade.java:78)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
The subid value of 2 exists if I run a native SQL query on the relevant parent table in the DB. Also the type of the subid is an int in the main entity class, the JPQL Facade class and the managed bean.
The JPQL:
public List<Itemdatapoint> chartType (int subid) {
Query q = em.createQuery("SELECT p FROM Itemdatapoint p JOIN p.series s WHERE s.master.item.subs.subid = :subid");
q.setParameter(subid, "subid");
return q.getResultList();
}
The managed bean:
#Named(value = "reportBean")
#SessionScoped
public class ReportBean implements Serializable {
#Inject
private ItemdatapointFacade facade;
public String typeSwitch1() {
subid = 2;
chartType = facade.chartType(subid);
for(Itemdatapoint e: chartType) {
status = e.getSeries().getMaster().getStatus();
if(status.equals("Include")) {
return "line";
}
}
return null;
}
The xhtml page:
<p:chart type="#{reportBean.typeSwitch1()}" model="#{reportBean.subLineChart1}"/>
I also tried a non Join JPQL just from a single table:
public List<Itemdatapoint> noJoin (int subid) {
Query q = em.createQuery("SELECT p FROM Itemdatapoint p WHERE p.pointid = :subid");
q.setParameter(subid, "subid");
return q.getResultList();
}
Similar problem:
java.lang.IllegalArgumentException: You have attempted to set a
parameter at position 2 which does not exist in this query string
SELECT p FROM Itemdatapoint p WHERE p.pointid = :subid.
I gather that IllegalArgumentException means that the selected entity does not exist or is not the correct type consistent with the query string in the facade class. But in my case the entity exists and the parameter is the correct type.
I would appreciate any help in understanding why i'm getting this error. Thank in advance!
UPDATE
Responding to the answer from lametaweb, I want to better understand the concept of JPA parameters.
According to the JPA documentation, the first argument of the setParameter method is the parameter name or number. The second argument is the object that should be bound to the named parameter. Why does the following work without throwing Illegal ArgumentException?
I tested an xhtml (web page):
<p:dataGrid id="rep1" columns="1" value="#{pageBean.itemPageList1}" var="items1" rows="4">
<p:commandLink value="#{items1.itemname}" action="#{pageBean.showItem1}" ajax="false"/>
</p:dataGrid>
The bean code:
public ListDataModel<Sectionitem> getItemPageList1() {
subid = 1;
reportStatus = "Include";
itemPageList1 = itemFacade.viewItems(subid, reportStatus);
return itemPageList1;
}
The JPA facade:
public ListDataModel<Sectionitem> viewItems(int subid, String stat) {
Query q = em.createQuery("select s from Sectionitem s JOIN s.subs c where c.subid = :subid AND s.status = :stat ORDER BY s.daterec");
q.setParameter("subid", subid);
q.setParameter("stat", stat);
ListDataModel<Sectionitem> res
= new ListDataModel<Sectionitem>(q.getResultList());
return res;
}
Why is it in this case, the object exist but in my original case the subid object does not exist?
You are invoking this method in your code:
setParameter(int position, Object value)
but you have to invoke this one instead:
setParameter(String name, Object value)
So your code should be:
q.setParameter("subid", Integer.valueOf(subid));
But, if you invoke:
q.setParameter(subid, "subid");
here the first parameter represents the position of the argument and the second the value for it. So you are passing a value of "subid" for the parameter in the second (2) position, which doesn't exist, because you only have one parameter in your JPQL query, hence the IllegalArgumentException exception.
Note: Why do you have a primitive type in your entity? Why not an Integer instead an int?
I'm new to JPA . I'm using spring data and I want to create an entity "File" (a folder is a File ) . each File can have a parent (or not).
this is my table's columns :id ,name,path ,mtime ,parentid.
I used this in my entities
#ManyToOne
#JoinColumn(name = "parentid")
private File parent;
public File() {}
public File(String name, String path,File parent) {
this.name = name;
this.path=path;
this.parent=parent;
}
#Override
public String toString() {
return String.format(
"File[ id=%d,name='%s', path='%s',parentid=%d]",
id,name, path,parent.id);
}
in my test I did this :
fileRepository.save(new File("file1","file1",null));
File file1=fileRepository.findOne((long) 1);
fileRepository.save(new File("file2","file2",file1));
and it inserted the first line with parentid NULL and a second line with parentid 1 (the first file).( I confirmed that on phpmyadmin)
I wanted to show the lines so i did this :
for (Object element : Lists.newArrayList(fileRepository.findAll())) {
System.out.println(element);
}
but it doesn't work .
when I removed parentid from my toString() function I get the correct result :
File[ id=1,name='file1', path='file1']
File[ id=2,name='file2', path='file2']
I get the same problem if I add a new column of type Long and when one of the lines have a NULL value in that column .
how Can I fix that ?
Your toString() method "doesn't work" because you're trying to get the id of a null parent, which obviously causes a NullPointerException. You need to test if the parent is null before trying to get and print its ID. Reading the message, type and stack trace of the exception would have allowed you to find this out. Using a debugger as well.
A variable of type long can't hold null. Primitive types are not nullable. So you need to use the type java.lang.Long for this field.
I am using the following JPA query and i am getting the java.lang.IllegalArgumentException: Cannot create TypedQuery for query with more than one return Exception.
TypedQuery<RaBdrRating> uQuery =
(TypedQuery<RaBdrRating>)entityManager.createQuery("
SELECT r.activePackage,SUM(r.duration),SUM(r.charge),COUNT(r)
FROM RaBdrRating r WHERE r.callType = :callType
and r.startDate between :startDate and :endDate
GROUP BY r.activePackage",RaBdrRating.class);
uQuery.setParameter("callType", model.getCallType());
uQuery.setParameter("startDate",startDate);
uQuery.setParameter("endDate",endDate);
List<RaBdrRating> listOfPackages = uQuery.getResultList();
Can any one tell me what is wrong in my query.....I am new to JPA and i am not getting what is the problem and strucked up here.If any one have idea please tell me.
I go the same error:
Cannot create TypedQuery for query with more than one return using requested result type
Solution:
Having a Query like this:
Select e.fieldA, e.fieldB, e.fieldC From Entity e
You have to declare a constructor with the parameters specified on query:
package somepackage;
public class Entity {
...
public class Entity() {}
public class Entity(Type fieldA, Type fieldB, Type fieldC) {
this.fieldA = fieldA;
this.fieldB = fieldB;
this.fieldC = fieldC;
}
....
}
Finally, modify your query
Select NEW somepackage.Entity(e.fieldA, e.fieldB, e.fieldC) From Entity e
You are indicating how the objectes will be created.
This seems to be this bug: https://hibernate.onjira.com/browse/HHH-6304
It is apparently fixed in version 4.1.5.