Reputation: 33
When I execute the following code the exception occurs:
Exception: org.springframework.orm.hibernate3.HibernateQueryException:
Not all named parameters have been set
Here is my code:
queryString = SET @quot=0,@latest=0,@comp='';
select B.* from (
select A.time,A.change,IF(@comp<>A.company,1,0) as LATEST,@comp:=A.company as company from (
select time,company,quote-@quot as `change`, @quot:=quote curr_quote
from stocks order by company,time) A
order by company,time desc) B where B.LATEST=1;
list = getHibernateTemplate().executeFind(new HibernateCallback(){
public Object doInHibernate(Session session)throws HibernateException,SQLException {
SQLQuery query = session.createSQLQuery(queryString);
query.setParameterList("list", custIds);
return query.list();
}
What is the reason for this behavior?
Upvotes: 3
Views: 1576
Reputation: 4498
you can create a named query and then use it in spring jpa repository or hibernate. This link helped me from similar problem.
Upvotes: 0
Reputation: 28146
It's a little bit hard to understand, what is exactly the query you are executing, but if you need to use the colon character in native query, in your case as "assign a value" operator, you should escape all the colon occurances with \\
in your java String with the query, so it could be like:
select B.* from (
select A.time,A.change,IF(@comp<>A.company,1,0) as LATEST,@comp\\:=A.company as company from (
select time,company,quote-@quot as `change`, @quot\\:=quote curr_quote
from stocks order by company,time) A
order by company,time desc) B where B.LATEST=1;
Update: seems, it is not possible yet to escape the colons in Hibernate native queries, there is an open issue about it. That means, that you are not able to use a colons in Hibernate native queries not for the named parameters. You can try to create a function and call it instead of calling a query.
Upvotes: 4