I have a domain class called Order and that class has hasMany relation with Item class.
When I am querying for the list of orders with certain restrictions I am getting as many instances of Order as there are items.
So for example Order instance has say references to 3 instances of Item then , criteria call on Order is returning 3 duplicate instances of Order. I am not sure but if it's worth mentioning that the domain class Order has fetchMode set to "eager".
I am really puzzled with what's going on there. Any help in this regard will be greatly appreciated. Snippet of code is attached:
def clazz = "cust.Order"
def criteria = clazz.createCriteria()
println("clazz == "+Order.list())// returning correct data i.e unique instance of order
def filter = {
// trimmed down all filtering criteria for debugging
}//close filter
List results = criteria.list(max:params?.max,offset:params?.offset,filter)
results.each{Object data->
println(data.getClass())
}
println("results == "+results)
Thanks again
One solution is to use this inside your query:
resultTransformer org.hibernate.Criteria.DISTINCT_ROOT_ENTITY
If you call criteria.listDistinct instead of criteria.list duplicates will be eliminated
Criteria API is just a wrapper for constructing a SQL query. In your case, the query in question has JOINs in it (because of the eager fetching), and returns a cartesian product of Orders and their matching Items. Each row returned is included in results as a separate Order instance.
The easiest way to remove duplicates is to put all the results in a Set, like this:
def resultSet = new HashSet()
resultSet.addAll(results)
println("results == " + resultSet)
You could also use dynamic finders, as in Order.findAllBy* .Depending on how complicated your filter is, this could be easy or tough :)
Related
I would like to order a collection first by priority and then due time like this:
#ods = Od.order(:priority, :due_date_time)
The problem is due_date_time is an instance method of Od, so I get
PG::UndefinedColumn: ERROR: column ods.due_date_time does not exist
I have tried the following, but it seems that by sorting and mapping ids, then finding them again with .where means the sort order is lost.
#ods = Od.where(id: (Od.all.sort {|a,b| a.due_date_time <=> b.due_date_time}.map(&:id))).order(:priority)
due_date_time calls a method from a child association:
def due_date_time
run.cut_off_time
end
run.cut_off_time is defined here:
def cut_off_time
(leave_date.beginning_of_day + route.cut_off_time_mins_since_midnight * 60)
end
I'm sure there is an easier way. Any help much appreciated! Thanks.
order from ActiveRecord similar to sort from ruby. So, Od.all.sort run iteration after the database query Od.all, run a new iteration map and then send a new database query. Also Od.all.sort has no sense because where select record when id included in ids but not searching a record for each id.
Easier do something like this:
Od.all.sort_by { |od| [od.priority, od.due_date_time] }
But that is a slow solution(ods table include 10k+ records). Prefer to save column to sort to the database. When that is not possible set logic to calculate due_date_time in a database query.
I was wondering if there is a way to do the following in a single query?
1) non_populated_models = PropertyPerson.select("property_people.id, count('items.recipient_person_id')").joins(:items).group('items.recipient_person_id, property_people.id')
2) populated_models = PropertyPerson.where(id: [non_populated_models])
Currently, the first group by query only returns the id, and count in the ProperyPerson object. Let's say there were 15 fields in the model and I didn't want to explicitly write them all out. Is there a way I can do this operation in a single query?
The join will work to limit the query to property_people with item and you you will get the extra column as an attr_reader.
people = PropertyPerson.select("property_people.*,
count('items.recipient_person_id') as items_count")
.joins(:items)
.group("property_people.id")
people.first.item_count
I have a domain object:
class Business {
String name
List subUnits
static hasMany = [
subUnits : SubUnit,
]
}
I want to get name and subUnits using HQL, but I get an error
Exception: org.springframework.orm.hibernate4.HibernateQueryException: not an entity
when using:
List businesses = Business.executeQuery("select business.name, business.subUnits from Business as business")
Is there a way I can get subUnits returned in the result query result as a List using HQL? When I use a left join, the query result is a flattened List that duplicates name. The actual query is more complicated - this is a simplified version, so I can't just use Business.list().
I thought I should add it as an answer, since I been doing this sort of thing for a while and a lot of knowledge that I can share with others:
As per suggestion from Yariash above:
This is forward walking through a domain object vs grabbing info as a flat list (map). There is expense involved when having an entire object then asking it to loop through and return many relations vs having it all in one contained list
#anonymous1 that sounds correct with left join - you can take a look at 'group by name' added to end of your query. Alternatively when you have all the results you can use businesses.groupBy{it.name} (this is a cool groovy feature} take a look at the output of the groupBy to understand what it has done to the
But If you are attempting to grab the entire object and map it back then actually the cost is still very hefty and is probably as costly as the suggestion by Yariash and possibly worse.
List businesses = Business.executeQuery("select new map(business.name as name, su.field1 as field1, su.field2 as field2) from Business b left join b.subUnits su ")
The above is really what you should be trying to do, left joining then grabbing each of the inner elements of the hasMany as part of your over all map you are returning within that list.
then when you have your results
def groupedBusinesses=businesses.groupBy{it.name} where name was the main object from the main class that has the hasMany relation.
If you then look at you will see each name has its own list
groupedBusinesses: [name1: [ [field1,field2,field3], [field1,field2,field3] ]
you can now do
groupedBusinesses.get(name) to get entire list for that hasMany relation.
Enable SQL logging for above hql query then compare it to
List businesses = Business.executeQuery("select new map(b.name as name, su as subUnits) from Business b left join b.subUnits su ")
What you will see is that the 2nd query will generate huge SQL queries to get the data since it attempts to map entire entry per row.
I have tested this theory and it always tends to be around an entire page full of query if not maybe multiple pages of SQL query created from within HQL compared to a few lines of query created by first example.
Hopefully this is a little clearer. I'm sorry but I'm very new to coding in general. I have multiple tables that I have to query in succession in order to get to the correct array that I need. The following logic for the query is as follows:
this gives me an array based upon the store :id
store = Stores.find(params[:id])
this gives me another array based upon the param .location found in the table store where that value equals the row ID in the table Departments
department = Departments.find(store.location)
I need to preform one last query but in order to do so I need to figure out which day of the meeting is needed. In order to do this I have to create the parameter day_of_meeting found in the table Stores. I try to call it from the array above and create a new variable. In the Table Departments, I there are params such as day_1, day_2 and so on. I need to be able to call something like department.day_1 or department.day_2. Thus, I'm trying to actually create the variable by join the words "department.day_" to the variable store.day_of_meeting which would equal some integer, creating department.day_1...
which_day = ["department.day_", store.day_of_meeting].join("")
This query finds uses the value found from the variable department.day_1 to query table Meeting to find the values in the corresponding row.
meeting = Meeting.find(which_day)
Does this make my problem any clearer to understand?
findmethod can only accept parameters like Meeting.find(1) or Meeting.find("1-xx").
so, what you need is Meeting.find(department.send("day_" + store.day_of_meeting.to_s))
Hope to help!
I am not sure if I am going about this the best way, but I will try to explain what I am trying to do.
I have the following domain classes
class User {
static hasMany = [goals: Goal]
}
So each User has a list of Goal objects. I want to be able to take an instance of User and return 5 Users with the highest number of matching Goal objects (with the instance) in their goals list.
Can someone kindly explain how I might go about doing this?
The easiest and most efficient way to achieve this is using plain SQL. Assuming you have these tables
users [id]
goals [id, description]
user_goals [user_id, goal_id]
You can have the following query to do what you need:
set #userId=123;
select user_id, count(*) as matched from user_goals
where user_id!=#userId
and goal_id in (select ug.goal_id from user_goals ug where ug.user_id=#userId)
group by user_id order by matched desc limit 5;
This takes a user id and returns a list of other users with matching goals, sorted by the number of matches. Wrap it up in a GoalService and you're done!
class GoalService {
def findUsersWithSimilarGoals(user) {
// ...
}
}
It may also be possible to do this with criteria or HQL, but with queries like this it's usually easier to use SQL.
If you're looking for a simple match, perhaps the easiest way would be to do a findAll for each Goal and then count the number of results that each other User appears in:
Map user2Count = [:]
for (goal in myUser.goals){
for (u in User.findAllByGoal(goal)){
def count = user2Count.containsKey(u) ? user2Count.get(u) : 0
count++
user2Count.put(u, count)
}
}
// get the top 5 users
def topUsers = user2Count.entrySet().sort({ it.value }).reverse()[0..5]
This may be too slow, depending on your needs, but it is simple. If many users share the same goals then you could cache the results of findAllByGoal.