如何从Drools中的列表中获取最大最小项 [英] How to get max min item from a list in Drools

问题描述

我有一个班级

class Person {
    public Date dateOfBirth;
    public List<Person> children;
}

我想创建一条Drools规则，该规则可以让我长大。例如：

I would like to create a Drools rule that gets me the oldest child. For example:

rule "Oldest Child"
    when
        $person: Person()
        $oldestChild: Person() from $person.children
    then
        insert($oldestChild)
end

按照书面规定，$ oldestChild是一个列表，但我真的很想成为一个实际的最大的孩子（单个对象而不是列表）。我玩了一点儿积累，但无法正常工作。有想法吗？

As written, $oldestChild is a list but I'd really like to be an actual oldest child (single object instead of a list). I played around with accumulate a bit but couldn't get it to work. Any ideas?

推荐答案

使用内联自定义代码进行累积会产生最老的孩子：

An accumulate with inline custom code produces the oldest child:

rule "oldest child"
when
  Person($pn: name, $pd: dateOfBirth, $children: children)
  Person($ocn: name) from accumulate(
    $child: Person( $cd: dateOfBirth) from $children,        
    init( Person minp = null; Date mind = new Date(); ),
    action( if( $cd.compareTo( mind ) < 0 ){
                minp = $child;
                mind = $cd;
            } ),
    result( minp ) )
then
  System.out.println( $pn + "'s oldest child is " + $ocn );
end

如果需要，可以实现自己的累加函数（在Java中）认真的工作-这是更多的工作，但却是更清洁的解决方案。请参阅文档。

You may implement your own accumulate function (in Java) if you need this for serious work - it is more work but the "cleaner" solution. See the docs.

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