I am experiencing various things while studying JPA, but I am too unfamiliar with it, so I would like to get some advice.
The parts I got stuck in during my study were grouped into three main categories. Could you please take a look at the code below?
#Repository
public interface TestRepository extends JpaRepository<TestEntity,Long> {
#Query(" SELECT
, A.test1
, A.test2
, B.test1
, B.test2
FROM TEST_TABLE1 A
LEFT JOIN TEST_TABLE2 B
ON A.test_no = B.test_no
WHERE A.test3 = ?1 # Here's the first question
if(VO.test4 is not null) AND B.test4 = ?2") # Here's the second question
List<Object[] # Here's the third question> getTestList(VO);
}
First, is it possible to extract test3 from the VO received when using native sql?
Usually, String test1 is used like this, but I wonder if there is any other way other than this.
Second, if extracting is possible in VO, can you add a query in #QUERY depending on whether Test4 is valued or not?
Thirdly, if I use List<Object[]>, can the result of executing a query that is not in the already created entity (eg, test1 in TEST_TABLE2, which is not in the entity of TEST_TABLE1) can be included?,
First, is it possible to extract test3 from the VO received when using native sql? Usually, String test1 is used like this, but I wonder if there is any other way other than this.
Yes, it is possible.
You must use, eg where :#{[0].test3} is equals vo.test3
[0] is position the first param, past for method annotated with #Query
#Query(value = "SELECT a.test1, a.test2, b.test1, b.test2
FROM test_table1 a
LEFT JOIN test_table2 b ON a.test_no = b.test_no
WHERE a.test3 = :#{[0].test3}", nativeQuery = true)
List<Object[]> getList(VO);
Second, if extracting is possible in VO, can you add a query in #QUERY depending on whether Test4 is valued or not?
You can use a trick eg:
SELECT ... FROM table a
LEFT JOIN table b ON a.id = b.id
WHERE a.test3 = :#{[0].test3}
AND (:#{[0].test4} IS NOT NULL AND b.test4 = :#{[0].test4})
Thirdly, if I use List<Object[]>, can the result of executing a query that is not in the already created entity (eg, test1 in TEST_TABLE2, which is not in the entity of TEST_TABLE1) can be included?
Sorry, but I not understand the third question.
Maybe this tutorial will help you: https://www.baeldung.com/jpa-queries-custom-result-with-aggregation-functions
Related
I'm using spring data JPA along with Hibernate. I have to find all entries by id, however I'm selecting only some of the columns.
I managed to do this by using specified constructor
#Query("SELECT new Foo(f.field1, f.field2, f.field3)
FROM FooTable f WHERE f.field1 = :field")
I need to make field1 as a DISTINCT, however putting this column into DISTINCT(field1) doesn't work.
Is there a way to make it work?
Try this:
#Query("SELECT new Foo(f.field1, f.field2, f.field3)
FROM FooTable f, FooTable f2
WHERE f.field1 = :field
AND f.field1 != f2.field1")
Note that it'll miss the null values.
I'm using Flask-SQLAlchemy with PostgreSQL. I have the following two models:
class Course(db.Model):
id = db.Column(db.Integer, primary_key = True )
course_name =db.Column(db.String(120))
course_description = db.Column(db.Text)
course_reviews = db.relationship('Review', backref ='course', lazy ='dynamic')
class Review(db.Model):
__table_args__ = ( db.UniqueConstraint('course_id', 'user_id'), { } )
id = db.Column(db.Integer, primary_key = True )
review_date = db.Column(db.DateTime)#default=db.func.now()
review_comment = db.Column(db.Text)
rating = db.Column(db.SmallInteger)
course_id = db.Column(db.Integer, db.ForeignKey('course.id') )
user_id = db.Column(db.Integer, db.ForeignKey('user.id') )
I want to select the courses that are most reviewed starting with at least two reviews. The following SQLAlchemy query worked fine with SQlite:
most_rated_courses = db.session.query(models.Review, func.count(models.Review.course_id)).group_by(models.Review.course_id).\
having(func.count(models.Review.course_id) >1) \ .order_by(func.count(models.Review.course_id).desc()).all()
But when I switched to PostgreSQL in production it gives me the following error:
ProgrammingError: (ProgrammingError) column "review.id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT review.id AS review_id, review.review_date AS review_...
^
'SELECT review.id AS review_id, review.review_date AS review_review_date, review.review_comment AS review_review_comment, review.rating AS review_rating, review.course_id AS review_course_id, review.user_id AS review_user_id, count(review.course_id) AS count_1 \nFROM review GROUP BY review.course_id \nHAVING count(review.course_id) > %(count_2)s ORDER BY count(review.course_id) DESC' {'count_2': 1}
I tried to fix the query by adding models.Review in the GROUP BY clause but it did not work:
most_rated_courses = db.session.query(models.Review, func.count(models.Review.course_id)).group_by(models.Review.course_id).\
having(func.count(models.Review.course_id) >1) \.order_by(func.count(models.Review.course_id).desc()).all()
Can anyone please help me with this issue. Thanks a lot
SQLite and MySQL both have the behavior that they allow a query that has aggregates (like count()) without applying GROUP BY to all other columns - which in terms of standard SQL is invalid, because if more than one row is present in that aggregated group, it has to pick the first one it sees for return, which is essentially random.
So your query for Review basically returns to you the first "Review" row for each distinct course id - like for course id 3, if you had seven "Review" rows, it's just choosing an essentially random "Review" row within the group of "course_id=3". I gather the answer you really want, "Course", is available here because you can take that semi-randomly selected Review object and just call ".course" on it, giving you the correct Course, but this is a backwards way to go.
But once you get on a proper database like Postgresql you need to use correct SQL. The data you need from the "review" table is just the course_id and the count, nothing else, so query just for that (first assume we don't actually need to display the counts, that's in a minute):
most_rated_course_ids = session.query(
Review.course_id,
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
order_by(func.count(Review.course_id).desc()).\
all()
but that's not your Course object - you want to take that list of ids and apply it to the course table. We first need to keep our list of course ids as a SQL construct, instead of loading the data - that is, turn it into a derived table by converting the query into a subquery (change the word .all() to .subquery()):
most_rated_course_id_subquery = session.query(
Review.course_id,
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
order_by(func.count(Review.course_id).desc()).\
subquery()
one simple way to link that to Course is to use an IN:
courses = session.query(Course).filter(
Course.id.in_(most_rated_course_id_subquery)).all()
but that's essentially going to throw away the "ORDER BY" you're looking for and also doesn't give us any nice way of actually reporting on those counts along with the course results. We need to have that count along with our Course so that we can report it and also order by it. For this we use a JOIN from the "course" table to our derived table. SQLAlchemy is smart enough to know to join on the "course_id" foreign key if we just call join():
courses = session.query(Course).join(most_rated_course_id_subquery).all()
then to get at the count, we need to add that to the columns returned by our subquery along with a label so we can refer to it:
most_rated_course_id_subquery = session.query(
Review.course_id,
func.count(Review.course_id).label("count")
).\
group_by(Review.course_id).\
having(func.count(Review.course_id) > 1).\
subquery()
courses = session.query(
Course, most_rated_course_id_subquery.c.count
).join(
most_rated_course_id_subquery
).order_by(
most_rated_course_id_subquery.c.count.desc()
).all()
A great article I like to point out to people about GROUP BY and this kind of query is SQL GROUP BY techniques which points out the common need for the "select from A join to (subquery of B with aggregate/GROUP BY)" pattern.
I am looking at some EF examples and trying to decipher what 'Query Projection' exactly equates to when doing LINQ to Entities or EntitySQL. I believe it is when the query results are filtered and projected into an anonymous type but not 100% sure.
Can someone please define this and maybe provide a small L2E query that uses an example of it?
Projection is when the result of a query is output to a different type than the one queried. Another article defined it as : the process of transforming the results of a query
Projection can be to an anonymous type, but could also be to a concrete type. If you come from a SQL world, it is akin to the columns listed in your SELECT clause.
Example selecting a sub-set of an object into an concrete type:
ParentObj.Select(x=> new ParentSlim { ParentID = x.ParentID, Name = x.Name } );
.
Example merging to object into a 3rd anonymous type:
Note: the select new portion is the projection.
from P in ParentObj.AsQueryable()
join C in ChildObj.AsQueryable() on P.ParentID == C.ParentID
select new { // <-- look ma, i'm projecting!
ParentID = P.ParentID,
Name = P.Name,
SubName = C.Name
RandomDate = DateTime.UtcNow()
}
I have the following question/problem:
I'm using JPQL (JPA 2.0 and eclipselink) and I wanna create a query that gives me the results sorted the following way:
At first the results sorted ascending by the best matches. After that should appear the inferior matches.
My objects are based on a simple class called 'Person' with the attributes:
{String Id,
String forename,
String name}
For example if I'm searching for "Picol" the result should look like:
[{129, Picol, Newman}, {23, Johnny, Picol},{454, Picolori, Newta}, {4774, Picolatus, Larimus}...]
PS: I already thought about using two queries, the first is searching with "equals" and the second with "like", although I'm not quite sure how to connect both queryresults...?
Hope for your help and thanks in advance,
Florian
If, as your question seem to imply, you only have two groups (first group : forename or name equals searched string; second group : forename or name contains searched string), and if all the persons of a given group have the same "match score", then using two queries is indeed a good solution.
First query :
select p from Person p where p.foreName = :param or p.name = :param
Second query :
select p from Person p where (p.foreName like :paramSurroundedWithPercent
or p.name like :paramSurroundedWithPercent)
and p.foreName != :param
and p.name != :param
Execute both queries (each returning a List<Person>), and add all the elements of the second list to the first one (using the addAll() method)
I have a query with a self join that looks like this,
select t1., t2. from table t1
left outer join table t2 on t2.LFT < t1.LFT
and t2.RGT > t1.RGT
AND t2.REG_CODE_PAR = 'ALL'
AND t1.STATUS_CODE = 'A'
AND t2.STATUS_CODE = 'A'
I'm using #NamedNativeQuery with a result set mapping to get the result.
#NamedNativeQuery(
name="findTree",
query="..... the query above",
resultSetMapping = "regionT")
With the following result set mapping
#SqlResultSetMapping(name = "regionT" , entities ={
#EntityResult(
entityClass = Tree.class
fields = {
#FieldResult(name = "regCode", column = "REG_CODE")
#FieldResult(name = "rgt", column = "RGT"),
#FieldResult(name = "lft", column = "LFT"),
#FieldResult(name = "name", column = "NAME"),
#FieldResult(name = "regCodePar", column = "REG_CODE_PAR"),
#FieldResult(name = "statusCode", column = "STATUS_CODE")
}
),
#EntityResult(
entityClass = TreeSelf.class
fields = {
#FieldResult(name = "regCode1", column = "REG_CODE")
#FieldResult(name = "rgt1", column = "RGT"),
#FieldResult(name = "lft1", column = "LFT"),
#FieldResult(name = "name1", column = "NAME"),
#FieldResult(name = "regCodePar1", column = "REG_CODE_PAR"),
#FieldResult(name = "statusCode1", column = "STATUS_CODE")
}
)
})
The entity class contains looks like this.
#NamedNativeQuery(...)
#SqlResultSetMapping(...)
#Entity
#Table(name = "table")
public class Tree implements Serializable {
#Id
#Column(name = "REG_CODE")
private String regCode; ... ..getters and setters...}
When I run the query using em.createQuery("findTree"), I get the exact same object in both
the 1st and 2nd elements of the returned object array.
Even if I create a class called TreeSelf that is identical to Tree and use it as the 2nd
EntityResult instead of having 2 EntityResults using the same entityClass, I get the same
result.
Can someone point out what's wrong with the configuration?
Let's see if I understand your question. You're expecting to capture two Tree entities from each native query result row. The first entity should be formed from t1's columns. The second entity should be formed from t2's columns. Contrary to expectation, you actually receive two instances formed from t1. No instances from t2 appear. You made a doppelganger Entity for Tree called TreeSelf while debugging, but TreeSelf is ultimately unnecessary and you want to get rid of it. Stop me if any of that was wrong.
I think the problem is due to ambiguous column names. Each column name in the native query appears twice, once from t1 and once from t2. The result mapper seems to be arbitrarily picking the first occurrence of each ambiguous column name for both Tree entities. I'm surprised that works at all. I would have expected an SQLException complaining about column reference ambiguity.
Also, are you sure you want a left outer join? What if no match is found for a t1 row? It will be paired with all NULL in t2's columns. Then you have a null-valued Tree entity. I think. I don't even know what the result mapper would do in that case. Perhaps you want an inner join?
Consider translating this native query into a JPQL query. (JPA Criteria API is just as well, but I find it more cumbersome for examples.) Here's a JPQL version of the native query:
SELECT t1, t2
FROM Tree t1, Tree t2
WHERE t2.lft < t1.lft AND t2.rgt > t1.rgt AND t2.regCodePar = 'ALL' AND
t1.statusCode = 'A' AND t2.statusCode = 'A'
N.B.: This changes the join semantics to inner instead of left outer.
Here's a sketch of code that could run this query:
EntityManager em = ... // EntityManager by injection, EntityManagerFactory, etc.
String jpql = ... // Like example above
TypedQuery<Object[]> q = em.createQuery(jpql, Object[].class);
for (Object[] oa : q.getResultList()) {
Tree t1 = (Tree)oa[0];
Tree t2 = (Tree)oa[1];
}
In case you are stuck with the native query for whatever reason, here's how you can work around the column name ambiguity. Instead of starting the native query like select t1.*, t2.*, alias each column with AS. The SELECT clause would resemble this:
SELECT t1.REG_CODE AS t1_REG_CODE, t1.RGT AS t1_RGT, (... rest of t1 cols ...),
t2.REG_CODE AS t2_REG_CODE, t2.RGT AS t2_RGT, (... rest of t2 cols ...)
The column attribute in each FieldResult must change accordingly. So the column attributes under the first EntityResult should all start with t1_ and the second's should all start with t2_.
I'd humbly recommend deleting the native query and sql result mapper and using JPA Query Language or Criteria API, if you can find a way.
Update: As confirmed in your comments, a useful answer to your question must preserve left (outer) join semantics. Unfortunately, JPQL and the Criteria API don't support complex left join conditions. There is no way to qualify a JPQL left join with an explicit ON condition.
To my knowledege, the only way to do a left outer join under the spec is by traversing an entity relationship. The JPA implementation then generates an ON condition that tests identity equality. The relevant spec bits are 4.4.5 "Joins" and 4.4.5.2 "Left Outer Joins".
To satisfy this constraint, each Tree you want to left-join to its ultimate parent must have an additional column storing the ultimate parent's id. You might be able to cheat around this constraint in a variety of ways (views?). But the path of least resistance seems to be modifying the native query to use aliased arguments, deleting TreeSelf, and updating the result mapper accordingly. Cleverer solutions welcome, though...