Since limit is not supported in #query of spring, we need alternative way for this other than pagination query param
#Query(" select something so where so.id>50 order by so.created desc")
#QueryHints(#QueryHint(name = "org.hibernate.annotations.FETCH_SIZE" , value = "1"))
optionam<something> findlatestSomething();
#Query(" select something s where s.id = select max(so.id) something so order by so.id desc")
optionam<something> findlatestSomething();
Related
I try to get orders of user by query :
#Query(value = "SELECT * FROM ORDERS WHERE USER_ID = ?1 AND CAST(CREATE_AT AS NVARCHAR(100)) LIKE ?2 OR CAST(GRAND_TOTAL AS NVARCHAR(100)) LIKE ?2 OR CAST(STATUS AS NVARCHAR(100)) LIKE ?2" , nativeQuery = true)
Page<Order> getOrdersByUserSearch(int userID, String searchS, Pageable pageable);
But it always return empty list. i run this code in SQL server and it work (?1 =2. ?2 = '2021-06-26').
If I try to change "NOT LIKE" instead of "LIKE" It run.
I dont want using query ( not native), Named query or specification method because it get more error.
Any advice?.
SQL like query requires % along with the value for a match. In case of ordered parameters in queries we can use:
#Query("SELECT m FROM Movie m WHERE m.rating LIKE ?1%")
List<Movie> searchByRatingStartsWith(String rating);
Click here for more info.
In your case the query string should be like this:
SELECT * FROM ORDERS WHERE USER_ID = ?1 AND CAST(CREATE_AT AS NVARCHAR(100)) LIKE %?2% OR CAST(GRAND_TOTAL AS NVARCHAR(100)) LIKE %?2% OR CAST(STATUS AS NVARCHAR(100)) LIKE %?2%
Using JPA specification classes or predicate builder. How can I convert this WHERE clause?
I am using an oracle db.
WHERE (SELECT listagg(reject_cd,':') within group (order by order_no) as rejectList
FROM REJECT_TABLE WHERE ID = transactio0_ id group by id) like '%06%'
The LISTAGG function is highly specific to Oracle, and is not supported by JPQL. However, you can still use a native query here, e.g.
#Query(
value = "SELECT ... WHERE (SELECT LISTAGG(reject_cd,':') WITHIN GROUP (ORDER BY order_no) AS rejectList FROM REJECT_TABLE WHERE ID = transactio0_ id GROUP BY id) LIKE '%06%'"
nativeQuery = true)
Collection<SomeEntity> findAllEntitiesNative();
Another option here might be to find a way to avoid needing to use LISTAGG. But, we would need to see the full query along with sample data to better understand your requirement.
Query:
select *
from easquestionsinfo
where questionname in(select questionname
from easresponseinfo
where isconflict = 'yes')
This query works fine and returns me the records from table 'easquestioninfo' when questionname is equal to the one returned by the inner query which returns set of questionname where isconflict='yes'.
JPA supports JPQL, SQL, and Criteria.
You can execute this SQL directly using createNativeQuery().
For JPQL, it depends on your object model, perhaps something like,
Select q fom QuestionInfo q where q.name in (Select r.name from ResponseInfo q2 where r.isConflict = 'yes')
See,
http://en.wikibooks.org/wiki/Java_Persistence/JPQL
I have this HQL where I need a subquery. I know it's not legal to make a subquery in order by, but I can't figure out how to do it
SELECT OBJECT(l) FROM InboundNotification l
INNER JOIN l.item item
WHERE l.job = ? ORDER BY (SELECT SUM(itemInst.qty)
FROM ItemInst itemInst
WHERE itemInst.receivedFromNotification_id = l.id) DESC, item.localId DESC
The above fails since I have the subquery in order by. How can I reconfigure it so this will work?
A sort in the Java code is not a option here even though it's almost as efficient.
ok, i haven't a notion of hql, but I'm gonna assume it's something like other query languages dive in here given that this question has remained unanswered for so long.
could you rewrite the query so it's something like this:
SELECT OBJECT(l), SUM(itemInst.qty) theSum
FROM InboundNotification l
INNER JOIN l.item item WHERE l.job = ?
INNER JOIN ItemInst on ItemInst.KEY = l.KEY
WHERE itemInst.receivedFromNotification_id = l.id)
GROUP BY OBJECT(l)
ORDER BY theSum
where ItemInst.KEY = l.KEY shows the appropriate relationship for your situation (if such a relationship exists)
I have a query similar to this:
SELECT username
FROM users
WHERE locationid IN
(SELECT locationid FROM locations WHERE countryid='$')
$ is a value I get from end user.
How could I run this query in CodeIgniter? I can't find a solution in CodeIgnite's user guide.
Thank you so much for your answers!
Regards!
Look here.
Basically you have to do bind params:
$sql = "SELECT username FROM users WHERE locationid IN (SELECT locationid FROM locations WHERE countryid=?)";
$this->db->query($sql, '__COUNTRY_NAME__');
But, like Mr.E said, use joins:
$sql = "select username from users inner join locations on users.locationid = locations.locationid where countryid = ?";
$this->db->query($sql, '__COUNTRY_NAME__');
Note that these solutions use the Code Igniter Active Records Class
This method uses sub queries like you wish but you should sanitize $countryId yourself!
$this->db->select('username')
->from('user')
->where('`locationId` in', '(select `locationId` from `locations` where `countryId` = '.$countryId.')', false)
->get();
Or this method would do it using joins and will sanitize the data for you (recommended)!
$this->db->select('username')
->from('users')
->join('locations', 'users.locationid = locations.locationid', 'inner')
->where('countryid', $countryId)
->get();
Also, to note - the Active Record Class also has a $this->db->where_in() method.
I think you can create a simple SQL query:
$sql="select username from user where id in (select id from idtables)";
$query=$this->db->query($sql);
and then you can use it normally.