Using Postgres with Play's Yabe Tutorial

Question

I'm trying to connect the Play's yabe tutorial with Postgres.

When i try to login as a particular user to post something in the Yabe app,

I get this following error,

JPAQueryException occured : Error while executing query from models.User where email = ? AND password = ?: ERROR: column user0_.id does not exist Position: 8

My User Model class is as below,

@Email
@Required
public String email;

@Required
public String password;
public String fullname;
public boolean isAdmin;

public User(String email, String password, String fullname) {
    this.email = email;
    this.password = password;
    this.fullname = fullname;
}

public static User connect(String email, String password) {
    return find("byEmailAndPassword", email, password).first();
    //return find("Select u from User u where u.email = ? and u.password = ?", email, password).first();
}

public String toString(){
    return email;
}

}`

I get this error in the connect method's "return find("byEmailAndPassword", email, password).first();" line.

I created the table in Postgres using the following query,

create table "User" ( id SERIAL, email varchar(255) NOT NULL, password varchar(255) NOT NULL, fullname varchar(255) NOT NULL, isAdmin boolean NOT NULL, primary key(id) );

I'm new to this framework and mvc style, and have no idea as to what the error is. Can anyone guide me as what should be done to fix this?

Thanks

Codemwnci · Accepted Answer · 2012-03-12 10:57:04Z

1

A few things here...

Are you extending the Model class, as specified in the YABE tutorial?
Why are you creating the table yourself. Play, via Hibernate will create the database table for you.

Try deleting the database table and let Play create it for you. If this still fails, then it can often be an issue with the Hibernate dialect for the database you are using. Make sure that you are using the Postgres Dialect in the application.conf file.

answered Mar 12, 2012 at 10:57

Codemwnci

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Comments

niels · Accepted Answer · 2012-03-12 15:28:57Z

1

I'm unsure if it was clever to create the table by your selves. The problem is that user is a protected word. It's much easier if you simple choose a different name for the table and let make hibernate do the rest for you. I choose

@Entity(name="YabeUser")
public class User extends Model {

and it works fine for me.

answered Mar 12, 2012 at 15:28

niels

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Code Lღver · Accepted Answer · 2012-03-14 05:40:52Z

0

Niels as the answer.

I'll add the following : DON'T USE CASE SENSITIVE NAME in your DDL.

In the same way don't use BLANKS in your DLL To summarize don't encapsulate your object's names (tables, columns, ...) between.

I've really seen a database designed by Business Analyst and all tables and columns names where really explicits : full explicit composite name with blanks. We had to have queries like this :

select "USER IDENTITY"."USER FIRSTNAME", "USER IDENTITY"."USER LASTNAME", "USER IDENTITY"."USER AGE" FROM "USER IDENTITY", "USER ROLE" WHERE "USER IDENTITY"."ROLE ID" = "USER ROLE"."ROLE ID" and "USER ROLE"."ROLE NAME" = 'ADMINISTRATOR'

It's a really a real case ! And don't want to see same thing again.

edited Mar 14, 2012 at 5:40

Code Lღver

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answered Mar 13, 2012 at 21:13

Javathought

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