我正在努力做下面的事情。 Plesse检查。
rule "new rule"
salience -101
dialect "mvel"
when
$pricingLineItem : PricingLineItem( $ackId : ackId, $prefix : prefix )
$baseUpChargeConfig : BaseUpChargeConfig( $baseOptionId : baseOptionId,
prefix == $prefix )
$pricingOptionType : PricingOptionType( ackId == $ackId,
$optionId : optionId, $optionValue : optionValue )
$baseOptionConfig : BaseOptionConfig( bOptionValue == $optionValue,
bOptionCode == $optionId ,id == $baseOptionId )
then
$pricingLineItem.increment($baseOptionId);
System.out.println("excuted - "+ $baseOptionId +" "+$baseOptionConfig);
end
一个PricngLineItem会有多个BaseUpChargeConfig对象匹配。在BaseUpChargeConfig对象中,我们获取所有相关的BaseOptionConfig对象,然后尝试使用PricingLineItem的PricingOptionType对象进行匹配。我需要使用与PricngLineItem的PricingOptionType对象最大匹配的最佳BaseUpChargeConfig对象。
修改
假设我有一个带有ackID,前缀值的PricingLineItem对象。 现在,我有一组基于PricingLineItem的前缀值的BaseUpChargeConfig对象。
现在在ackId值上,我在规则引擎中有一定数量的PricingOptionType对象。
并且在baseOptionId值上,我有多个BaseOptionConfig对象。
在PricingOptionType和BaseOptionConfig对象中,我需要比较optioncode和option值。
如果两者都匹配,我需要为perticuler BaseUpChrageConfig收集所有匹配的定价选项类型。
以同样的方式,这将检查所有其他BaseUpChrageConfig对象BaseOptionConfig并匹配。
现在匹配最高的BaseOptionConfig对象;我们将选择BaseUpChargeConfig作为我们目的的最佳对象。
我希望你能清楚。
目前我通过传递所有三个并在java中计算来完成java方法。
public void matchOptions(BaseUpChargeConfig config,List pricingOptionList, 列出baseOptionList){
if ((pricingOptionList != null && !pricingOptionList.isEmpty())
&& (baseOptionList != null && !baseOptionList.isEmpty())) {
List<PricingOptionType> matchedOption = null;
matchedOption = new ArrayList<PricingOptionType>();
for (PricingOptionType pOption : pricingOptionList) {
int matchCount = 0;
for (BaseOptionConfig bConfig : baseOptionList) {
boolean optioncodeMatch = pOption.getOptionCode() == bConfig.getBaseOptionCode();
boolean optionValueMatch = pOption.getOptionValue() == bConfig.getBaseOptionValue();
if (optioncodeMatch && optionValueMatch) {
matchedOption.add(pOption);
matchCount++;
}
}
if (matchCount > 0) {
if (bestBaseUpChargeConfig != null) {
optionMatchCount = matchCount;
bestBaseUpChargeConfig = config;
matchedPrcOptionList = matchedOption;
} else if (matchCount == optionMatchCount) {
bestBaseUpChargeConfig = null;
matchedOption = null;
matchedPrcOptionList.clear();
} else if (matchCount > optionMatchCount) {
optionMatchCount = matchCount;
bestBaseUpChargeConfig = config;
matchedPrcOptionList = matchedOption;
} else {
// do nothing
}
}
}
} else {
// do nothing
}
}
由于
答案 0 :(得分:1)
这可以用5.5编译,所以它也不应该是6.x的问题。
除非您考虑涉及派生事实的更复杂的评估,否则无法帮助重复累积。
rule "find best BaseUpChargeConfig"
when
// pick up some PricingLineItem
$pli: PricingLineItem( $prefix: prefix, $ackId : ackId )
// it should have a BaseUpChargeConfig with a matching prefix
$bucc: BaseUpChargeConfig( prefix == $prefix,
$baseOptionId : baseOptionId )
// count BaseOptionConfigs (linked to BaseUpChargeConfig) matching
// PricingOptionTypes, by option id/code and option value
accumulate(
BaseOptionConfig( id == $baseOptionId,
$ocod: bOptionCode, $oval: bOptionValue )
and
PricingOptionType( ackId == $ackId,
optionId == $ocod, optionValue == $oval );
$count: count(1) )
// The $count computed above is the maximum if we don't have another
// BaseUpChargeConfig (for that prefix) where the count of the
// subordinate BaseOptionConfigs is greater than $count
not(
( BaseUpChargeConfig( this != $bucc,
prefix == $prefix,
$baseOptionId2 : baseOptionId )
and
accumulate(
BaseOptionConfig( id == $baseOptionId2,
$ocod2: bOptionCode, $oval2: bOptionValue )
and
PricingOptionType( ackId == $ackId,
optionId == $ocod2, optionValue == $oval2);
$count2: count(1);
$count2 > $count ) ) )
then
System.out.println( "best BaseUpChargeConfig: " + $baseOptionId );
end