Defining default sort-order in Grails/GORM

问题

Let's say I have definied a User object using GORM. Each user can have zero or more Login:s. Each Login has a timestamp. When retrieving user.logins I want the logins to be sorted based on the value of login.date. What is the correct Grails way to achieve this?

Example: I want the following code to list all the user's logins in ascending order.

<g:each var="login" in="${user.logins}">
  <tr>
    <td>${login.date}</td>
  </tr>
</g:each>

These are the referenced classes:

class User {
  ...
  def hasMany = [logins: Login]
  static fetchMode = [logins: "eager"]
}

class Login {
  Date date
  ...
  def belongsTo = [User]
}

I'm running Grails 1.0.4 which is the latest stable release.

回答1:

They show how to do this on the GORM page in the reference guide (section 5). The bit you want is near the bottom of that document is the section you want. They have two simple examples:

class Airport {
    …
    static mapping = {
        sort "name"
    }
}

class Airport {
    …
    static mapping = {
        sort name:"desc"
    }
}

They also have an example of sorting on an association:

class Airport {
    …
    static hasMany = [flights:Flight]
    static mapping = {
        flights sort:'number'
    }
}

回答2:

Just make the Login Class implement the Comparable interface:

class Login implements Comparable {

    // ...

    Date date

    public int compareTo(def other) {
        return date <=> other?.date // <=> is the compareTo operator in groovy
    }

}

and declare the relation to be a SortedSet:

class User {
  ...
  def hasMany = [logins: Login]               
  SortedSet logins

  static fetchMode = [logins: "eager"]
}

回答3:

The handling of default sort order in Grails/GORM seems to have been radically simplified in Grails 1.1:

Grails 1.1 release notes (search for "Default Sort Order")

来源：https://stackoverflow.com/questions/520187/defining-default-sort-order-in-grails-gorm

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grails

gorm

gsp