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

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;

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", ...

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{

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.