org.hibernate.exception.SQLGrammarException: could not execute query?

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I got this exception when i try to get the list using NamedQuery:

org.hibernate.exception.SQLGrammarException: could not execute query

Below I mentioned code:

Entity Class Code:

@Table(name = "tbl_users")
@XmlRootElement
@NamedQueries({@NamedQuery(name = "TblUsers.findAll", query = "SELECT t FROM TblUsers t")});

DAO implement Code:

org.hibernate.Query query = session.getNamedQuery("TblUsers.findAll");
List list = query.list();

Please provide solution for this exception.

Was facing same issue for a while and figured out that the problem was due to the Table name being different from the class(or Entity) name in the database. Added the @Table(name = actual_table_name) annotation and it worked.

How To Solved : org.hibernate.exception.SQLGrammarException , SQLGrammarException: could not execute query". Post by: Irliyanti Request processing failed; nested exception is org.hibernate.exception. [Solved]org.hibernate.exception.SQLGrammarException: could not execute statement December 23, 2019 Saurabh Gupta Leave a comment “org.hibernate.exception.SQLGrammarException: could not execute statement” this exception generally occurred when there is some issue with generated query grammar or syntax.

Get the SQL query that Hibernate is generating (using hibernate.show_sql or, preferably, Hibernate's SQL logging), and execute it against the database yourself. That will most likely help steer you in the right direction.

org.hibernate.exception.SQLGrammarException: could , org.hibernate.exception.SQLGrammarException: could not execute query. Post by: anji anjaiah , Greenhorn. Mar 29, 2015 21:40:22. HI Team , Please help here​  Irliyanti Rahmadhani Lubis wrote:org.hibernate.exception.SQLGrammarException: could not execute query I'm trying to solved them, but it still not working. What happens when you execute that query directly on the database using a SQL Client program like Squirrel or the client which comes with your database? Kind regards, Roel

Try this one it could work. It Perfectly worked for me.

1) Class level annotation.

@NamedQuery(name="UserDetails.byId" , query="from UserDetails where userId = ?")

2) Get record using NamedQuery

Query qry2 = sf.getCurrentSession().getNamedQuery("USER_DETAILS.byName")    ;
        qry2.setString(0, "Angad Bansode");
        List<UserDetails> user = qry2.list();
        for (UserDetails userDetails : user) {
            System.out.println("User Details by named native query  name = " + userDetails.getUserName() + ", aadhaar no  = " + userDetails.getAadharNo());
        }

org.hibernate.exception.SQLGrammarException, I am getting Exception as: org.hibernate.exception.SQLGrammarException: could not execute query. The reason is: SEVERE: ORA-00904:  org.hibernate.exception.SQLGrammarException: could not execute query Caused by: java.sql.SQLException: ORA-00903: invalid table name

seems like this question is little old but any way once i added below line to hibernate config files it worked for me.

<property name="show_sql">true</property>

org.hibernate.exception.SQLGrammarException: could not prepare , SQLGrammarException: could not prepare statement - Table “USER_DETAIL” not found; USER_FIND_BY_USER_NAME, query = "SELECT u FROM UserDetail u WHERE u. I tried to execute login() method without JUnit and it's working. Hi All, I am getting Exception as: org.hibernate.exception.SQLGrammarException: could not execute query The reason is: SEVERE: ORA-00904: "THIS_"."COMPANYTYPE_COMPANY_TYPE_ID": invalid identifier

hibernate.exception.SQLGrammarException: could not execute query, SQLGrammarException: could not execute statement” this exception generally occurred when there is some issue with generated query grammar  By the way hibernate does not force you to create a new entity for many-to-many join tables (which don't have any more attribute but the foreign keys). But I believe it is a good practice to have an entity for that relationship, cause most of the times some attributes will be defined for the relation in future.

SQLGrammarException MySQLSyntaxErrorException, SQLGrammarException: could not execute query at org.hibernate.exception.​SQLStateConverter.convert(SQLStateConverter.java:67) Could someone please​  exception: org.hibernate.exception.SQLGrammarException: could not execute native bulk manipulation query. After looking in more than 25 google results I couldn't find any clue about what could be happening. But somehow I got illuminated and found what was happening. That's why I'm writing this post.

[Solved]org.hibernate.exception.SQLGrammarException: could not , org.hibernate.exception.SQLGrammarException: could not execute query. Published on October 26, 2013 by sqlwizard. General This problem is due to the use  [11/26/19 15:45:54:162 IST] 00000023 SystemErr R org.hibernate.exception.GenericJDBCException: could not insert: [com..databean.Incident336AttachmentBean] Vote Up 0 Vote Down 5 months ago

Comments
  • Have you tried just "from TblUsers t"? This is the correct HQL syntax.
  • @Olaf - "select t from TblUsers t" is correct HQL as well
  • You need to show the whole stack trace. SQLGrammarException means SQL (as translated by Hibernate) was not accepted by your database. Make sure that the table exists (in proper schema).
  • I've tried using hibernate.show_sql but all I get is insert into table (column) values (?) How can we actually see the REAL SQL and not just a bunch of ? marks?