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Criteria Query and pagination params

Wednesday, July 14th, 2010
Posted by Bhagwat Kumar

I have been using the following code to get paginated result and the total number of results returned irrespective of the pagination params.

 def result=SampleDomain.createCriteria().list(){
// multiple restrictions
   maxResults(params.max)
   firstResult(params.offset)
} // Return type is ArrayList
 
Integer  totalResult=SampleDomain.createCriteria().count(){
// multiple restrictions
// maxResults(params.max)
// firstResult(params.offset)
}

Clearly duplicating the same closure except for the pagination restrictions was not a good solution.

After a little googling and reading mailing lists I got the solution. Passing pagination params to createCriteria.list() returns result of type PagedResultList which provides many useful methods. The getTotalCount() method of PagedResultList class returns the actual number of results returned irrespective of the pagination restrictions(maxResults and firstResult). Also the result contains only those records fulfiling maxResults and firstResult restrictions.

def result=SampleDomain.createCriteria().list(max:params.max, offset:params.offset){
// multiple/complex restrictions
   maxResults(params.max)
   firstResult(params.offset)
} // Return type is PagedResultList

Thanks to all the active users of Grails mailing list.

Here are few useful links:
http://www.pubbs.net/grails/200912/2269
http://www.grails.org/doc/1.0.x/api/grails/orm/PagedResultList.html
http://jira.codehaus.org/browse/GRAILS-2672

Bhagwat Kumar
bhagwat(at)intelligrape(dot)com

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Grails criteria query example for unidirectional one-to-many relationships

Monday, June 14th, 2010
Posted by Aman Aggarwal

Following is an example of unidirectional one-to-many relationship:

class Employee {
    String name
    static hasMany = [roles: Role]
}
 
class Role {
    String name
}

How can we find all the employees with a particular role – “Grails Developer”?

Role role = Role.findByName("Grails Developer")

One way of doing this can be:

Employee.list().findAll{role in it.roles}

This is an inefficient way since, we are fetching all the records from the database every time. Also, in case we are showing paginated view of results, we will need to manage pagination ourself.

A better way of this will be using createCriteria() query:

List<Employee> employees = Employee.createCriteria().list{
    roles{
        eq('id', role.id)
    }	
    firstResult(20)
    maxResults(20)	
}

I am wondering what will be the query in case Role.groovy is an enum.

–
~Aman Aggarwal
aman@intelligrape.com

http://www.IntelliGrape.com/

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