public class Person {
List<PersonAddress> personAddress;
}
public enum AddressType {
CURRENT, PREVIOUS;
}
public class PersonAddress{
Address address;
AddressType type
Integer timeAtAddress;
}
public class Address {
String city;
String country;
String street;
}
我必须编写一些代码来验证drools中的PersonAddress。 规则1.如果person具有PersonAddress实例列表,如果其中一个是AddressType == CURRENT和timeAtAddress&lt; 3,然后我想查找该列表是否包含AddressType == PREVIOUS的地址。
规则2.如果以上条件为真,那么我想获取AddressAddress实例,其中AddressType == PREVIOUS,
Drools版本5.5.0.Final,
Java 1.7
可以使用功能
这是我尝试过的,但它不起作用
function boolean isPreviousAddressExist(java.util.List list) {
if(list.isEmpty()) {
return false;
}
boolean validRecordFound = false;
for(int addressIndex = 0; addressIndex < list.size(); addressIndex++) {
PersonAddress pa = (PersonAddress)list.get(addressIndex);
if(AddressType.CURRENT.equals(pa.getAddressType()) && pa.getTimeAtAddress() != null && pa.getTimeAtAddress() < 3) {
validRecordFound = true;
break;
}
}
boolean previousRecordFound = false;
if(validRecordFound) {
for(int addressIndex = 0; addressIndex < list.size(); addressIndex++) {
PersonAddress pa = (PersonAddress)list.get(addressIndex);
if(AddressType.PREVIOUS.equals(pa.getAddressType())) {
previousRecordFound = true;
break;
}
}
} else {
previousRecordFound = true;
}
return previousRecordFound;
}
rule "Previous-Physical-Home Address is required for Time at Current-Physical-Home"
when
$quotation:Quotation()
eval(!isPreviousAddressExist($quotation.getApplicantList()))
then
runningResults.addRunningResult(new BusinessRuleRunningResult(null, " A Previous physical Home Address is required.", false));
end
答案 0 :(得分:1)
以下是两条规则,使用extends
来避免重复:
rule "brief CURRENT"
when
Person( $name: name, $pa: personAddress )
PersonAddress( type == AddressType.CURRENT, timeAtAddress < 3 ) from $pa
then
end
rule "no PREVIOUS"
extends "brief CURRENT"
when
not PersonAddress( type == AddressType.PREVIOUS ) from $pa
then
System.out.println( "invalid. " + $name );
end
rule "has PREVIOUS"
extends "brief CURRENT"
when
$ppa: PersonAddress( type == AddressType.PREVIOUS ) from $pa
then
System.out.println( "valid. " + $name +
" at " + $ppa.getAddress().getCity() );
end
这假设每个人只有一个CURRENT;如果没有,规则可能不止一次。使用exist
来避免。
我没有尝试在您的代码中找到错误,因为&#34;没有工作&#34;太模糊了无论如何,如果你正在使用规则,编写程序代码进行检查的重点是什么?