Named parameter does not follow java identifier rules in MySQL/SQL/JPQL Query

Question

I would like to create a JPQL query that shows the list of users in the same group as the logged in user.

Indeed, that's what I already did

public List<Utilisateur> getListUser(Long id) {
    return getEntityManager()
            .createQuery("SELECT u FROM Utilisateur u WHERE u.group.id =:" +id)
            .getResultList();
}

and my User tables is as follows

User: id, name, tel, email, group, ...

Group: id, name, type.

My method to retrieve the group of the current user

public Long currentIdGroup() {
    Utilisateur usere = EntityRealm.getUser();
    if (usere == null) {
        return null;
    }
    return EntityRealm.getUser().getGroup().getId();
}

The declaration of the list in the Bean.

public List<Utilisateur> getListusersGroup() {
    return this.utilisateurService.getListUser(currentIdGroup());

}

But, I'm getting this error in glassfish

Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.JPQLException Exception Description: Syntax error parsing [SELECT u FROM Utilisateur u WHERE u.group.id =:1]. [51, 53] The named input parameter ''{0}'' is not following the rules for a Java identifier.

Thank you in advance for your crucial help.

Many things to you.

Remove the colon

egallardo
– egallardo

2018-01-24 21:25:14 +00:00
Commented Jan 24, 2018 at 21:25 — egallardo
– egallardo, Commented Jan 24, 2018 at 21:25

egallardo · Accepted Answer · 2018-01-24 21:25:49Z

3

Try removing the colon:

.createQuery("SELECT u FROM Utilisateur u WHERE u.group.id =" +id)

Ideally you should use:

.createQuery("SELECT u FROM Utilisateur u WHERE u.group.id =:param")

and then:

.setParameter("param", id);

to avoid SQL injection.

answered Jan 24, 2018 at 21:25

egallardo

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2 Comments

Rod. TCH Over a year ago

in fact I forgot to say that group is a foreign key. Like this User: id, name, tel, email, FK group_id, ... Group: id, name, type. Sorry for that.

Rod. TCH Over a year ago

Thanks egallardo for your help. It's work. Many thanks.

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