logo
down
shadow

nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet ,Spring4,Hibernate4


nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet ,Spring4,Hibernate4

Content Index :

nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet ,Spring4,Hibernate4
Tag : hibernate , By : johntynan
Date : November 23 2020, 01:01 AM

it should still fix some issue As I remember, MySQL table/column names are by default case sensitive on linux environment, so try updating User.hbm.xml accordingly.

Comments
No Comments Right Now !

Boards Message :
You Must Login Or Sign Up to Add Your Comments .

Share : facebook icon twitter icon

Error performing load command : org.hibernate.exception.SQLGrammarException: could not extract ResultSet Exception in th


Tag : java , By : Aki Björklund
Date : March 29 2020, 07:55 AM
hope this fix your issue The problem is with the entity Zamowienie you are mapping the field adres to a column named imie:
@Column(name="imie", length=50)
private String adres;
@Column(name="adres", length=50)
private String adres;

nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet ,Spring4,Hibernate4.please


Tag : hibernate , By : Rb.Ridge
Date : March 29 2020, 07:55 AM
it fixes the issue I think the problem here is that the table bsedb.client doesn't exist because as shown in your sceenshot the table name is:
bsedb.bse_client'
@Table(name="BSE_CLIENT",...
@Table(name="CLIENT", ...

nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet Hibernate+SpringMVC


Tag : java , By : protagonist
Date : March 29 2020, 07:55 AM
wish of those help I am trying to fetch records from the database and I am having the above error , You are missing couple of things
import javax.persistence.Cacheable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.persistence.Transient;

import org.codehaus.jackson.annotate.JsonProperty;
import org.codehaus.jackson.map.annotate.JsonSerialize;
import org.hibernate.annotations.Cache;
import org.hibernate.annotations.CacheConcurrencyStrategy;


import java.io.Serializable;
import java.util.Date;
@Entity
@Table(name="User")

public class User implements Serializable{

SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet


Tag : mysql , By : user176445
Date : November 26 2020, 06:28 AM
Does that help Whenever you face such type of error, while running Spring Boot + JPA CRUD Example, Just check once that you have defined ID field using AUTO_INCREMENT option or not. I was having such type of Error because I did not require ID field. BUT Spring Boot Hibernate checks for Unique A_I ID field.
If you don't require ID field, then use and Your problem will be solved.

org.hibernate.exception.SQLGrammarException: could not extract ResultSet


Tag : hibernate , By : Chris Hanley
Date : March 29 2020, 07:55 AM
Related Posts Related QUESTIONS :
shadow
Privacy Policy - Terms - Contact Us © scrbit.com