Grails - How to get list of distinct User Objects

Question

Is there a way that, i can get list of distinct User objects(based on username). And still get result as a List of User Objects rather than, List of username's.

My code is

def criteria = User.createCriteria()
def users = criteria.list() {
    projections {
        distinct("username")
    }
    setResultTransformer(CriteriaSpecification.ROOT_ENTITY)
}
return users

Currently am getting List of the usernames, not User.

Answer 1

Ya projection is like filtering and selecting by username you should change it to

def criteria = User.createCriteria()
def users = criteria.listDistinct() {
    projections {
        groupProperty("username")
    }
}
return users

JOB DONE!

Answer 2

One of these should work - I haven't tested any of them, I leave that up to you :)

• User.list().unique()
• User.list().unique() with the equals() method on the User domain class overridden to compare objects using the username
• User.list().unique { it.username } (might need toArray() after list())

Answer 3

def criteria = User.createCriteria()
def users = criteria.list() {
    projections {
        distinct("username")
    }
    setResultTransformer(CriteriaSpecification.ROOT_ENTITY)
}

Just replace setResultTransformer(CriteriaSpecification.ROOT_ENTITY) with resultTransformer(ALIAS_TO_ENTITY_MAP). You will get a list of string as a result

otherwise just replace .list with .listDistinct and use do not need distinct("username"), just can be property("username");

Usually people get problems with pagination. not results. If you already had something like:

User.createCriteria().list([max:params.max,offset:params.offset],{
createAlias("others", "others", CriteriaSpecification.LEFT_JOIN);

ilike("others.firstName", "%${query}%");
});

It could result in row duplicates. Because .listDistinct() does not support pagination, just add

resultTransformer(CriteriaSpecification.DISTINCT_ROOT_ENTITY);

So query will look like this:

User.createCriteria().list([max:params.max,offset:params.offset],{
    resultTransformer(CriteriaSpecification.DISTINCT_ROOT_ENTITY);
    createAlias("others", "others", CriteriaSpecification.LEFT_JOIN);

    ilike("others.firstName", "%${query}%");
    });

Answer 4

Where ever you got a collection (list, array, ...) (I don't know if work with any type of collection, but work in all that i could test). Use unique{ it.property }:

Example:

---

def users = []
for (def room in rooms) {
    users.addAll(room.users)
}
return users.unique{ it.id }