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

Question

I use play framework !! But when I run my project it give me this

who can help me ?

this is my model:

package models;
import java.util.*;
import javax.persistence.*;
import play.db.jpa.*;
import play.db.jpa.Model;

@Entity
@Table(name="GxkAccount")
public class GxkAccount extends Model {

    private String Account;
    private String Psw;

    public String getAccount() {
        return Account;
    }
    public void setAccount(String account) {
        Account = account;
    }
    public String getPsw() {
        return Psw;
    }
    public void setPsw(String psw) {
        Psw = psw;
    }

    public static List<GxkAccount> GetList()
    {
        List<GxkAccount> infoList=GxkAccount.findAll();
        return infoList;
    }


}

Please share the full exception message. And if you can have the actual query generated by Hibernate, that would be great. — Adeel Ansari
– Adeel Ansari, Commented Nov 3, 2010 at 3:42

Adeel Ansari · Accepted Answer · 2010-11-03 03:48:58Z

4

You are completely missing the mapping annotations for the properties of your class.

P.S. Please try to follow the Java naming conventions

edited Nov 3, 2010 at 3:48

answered Nov 3, 2010 at 3:43

Adeel Ansari

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1 Comment

Adeel Ansari Over a year ago

@niels: Then the generated query is essentially needed.

sth · Accepted Answer · 2015-01-25 15:45:31Z

1

Using mysql, we also faced this type of issue. We found in play framework application.conf:

jpa.dialect=org.hibernate.dialect.PostgreSQLDialect

we replaced this with

jpa.dialect=org.hibernate.dialect.MySqlDialect.

This solved the problem. If you are facing this issue you can try out this configuration setting.

edited Jan 25, 2015 at 15:45

sth

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answered Nov 17, 2011 at 7:08

Mohan Shanmugam

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Comments

sth · Accepted Answer · 2015-01-25 15:40:36Z

We also faced the same issue. We were having create in the xml and @GeneratedValue on the id column. The resolution is remove the @GeneratedValue annotation and put the value of the id manually, also the jpa takes long by default so give long value e.g 1l.

To do the auto generation follow some another rule.

The issue around the JPA related auto generated Id is resolved as below:

Modify the Person.java model class to have the following annotations for the Id attribute:

@Id
@TableGenerator(name="TABLE_GEN",table="T_GENERATOR",pkColumnName="GEN_KEY",pkColumnValue="TEST",valueColumnName="GEN_VALUE",initialValue=1,allocationSize=1)
@GeneratedValue(strategy=GenerationType.TABLE, generator="TABLE_GEN")
public Long Id;

This will create a table in the mysql schema called T_GNERATOR which will have the tracking of the next value for Id and JPA over hibernate knows how to retrieve this value. The assumption is that the initial value for the Id is 1 and it is incremented by 1 on each new insertion into it as is obvious from the attributes of the annotation.

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