Question

我想知道是否可以使用drools accumulate expression获取具有更高优先级的对象？这是我的代码：

rule "dropShiftWithTheLowestPriorityTaskForWeekdayMorningShiftReassignment"
when 
    ShiftAssignment( isNeedToReassign == true, shiftType.code == 'E', weekend == false,  
        $shiftDate : shiftDate, 
        $shiftType : shiftType )

    $max : Number(intValue >= 0) from accumulate(
        $assignment : ShiftAssignment(
            weekend == false,
            shiftDate == $shiftDate,
            shiftType == $shiftType, 
            $totalWeight : totalWeight),
        max($totalWeight)
    )
then
    System.out.println('----------');
    System.out.println('max='+$max);

我只获得了最大的totalWeight，但我不知道如何获得包含该totalWeight的对象。请帮帮我，谢谢。

Answer 1

几天前发布了一个类似的问题： How to get max min item from a list in Drools

有两个解决方案：

创建自己的累积功能
使用2个简单模式

按照第一种方法，您需要编写如下内容：

when
    $maxAssignment: ShiftAssignment() from accumulate(
        $sa: ShiftAssignment(),        
        init( ShiftAssignment max = null; ),
        action( if( max == null || max.totalWeight < $sa.totalWeight ){
            max = $sa;
        } ),
        result( max ) )
then
    //$maxAssignment contains the ShiftAssignment with the highest weight. 
end

请注意，此实现仅应用于原型设计或测试目的。在Java中实现自定义累积函数被认为是一种很好的做法。

按照第二种方法，您的规则可以重写为：

when 
    $maxAssignment: ShiftAssignment(..., $maxWeight: totalWeight)
    not ShiftAssignment(..., totalWeight > $maxWeight)
then
    //$maxAssignment contains the ShiftAssignment with the highest weight. 
end

希望它有所帮助，

Drools获得具有更高优先级的对象

1 个答案: