Grails - How to get list of distinct User Objects

问题:

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.

回答1:

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

def criteria = User.createCriteria() def users = criteria.listDistinct() {     projections {         groupbyproperty("username")     }    // setResultTransformer(CriteriaSpecification.ROOT_ENTITY) } return users 

JOB DONE!

回答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())

回答3:

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 } 

回答4:

This will get you the distinct user object based on userName

def userInstance = User.list().unique{ it.user_name}

回答5:

Using where query(Detached criteria):

def userListQuery = User.where{ // Your search criteria }  def userList = userListQuery.list().unique{ it.username } 

it should result one query for distinct results.

回答6:

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}%");     }); 

文章来源: Grails - How to get list of distinct User Objects