Question

在流口水中，当我执行具有事实的无状态会话时，我们可以选择找到规则中所有条件元组都满足的条件。

ex：如果规则RUL1中有规则条件， car.schi =='A'|| car.carKind str [startsWith]'A'

如果RUL1因car.schi而感到满意，那么我们是否有任何API可以获取此信息（car.schi）。

我给出的示例比较简单，但是我们的实际业务规则非常复杂，如下所示，

（car.carKind！= \“ EZZ \” && car.carKind！= \“ ENG \” && car.carKind！= \“ ETD \”）&&（（car.schi包含\“ N1 \” | | | car.schi包含\“ N2 \”）||（（（car.schi包含\“ IH \” || car.schi包含\“ N4 \” || car.schi包含\“ OM \” ||汽车。 schi包含\“ DA \” || car.schi包含\“ N5 \” || car.schi包含\“ PA \” || car.schi包含\“ FG \” || car.schi包含\“ PL \ “ || car.schi包含\” PC \“ || car.schi包含\” PO \“ || car.schi包含\” NG \“ || car.schi包含\” OX \“ || car.schi包含\“ OP \” || car.schi包含\“ NS \” || car.schi包含\“ FS \” || car.schi包含\“ FL \” || car.schi包含\“ N3 \” || car.schi包含\“ CM \” || car.schi包含\“ DW \” || car.schi包含\“ PB \”）&&（validateElementRule（$ trainrulesRequestDTO.getElementRuleMap（），\“ 1_N \” ，true，$ trainrulesRequestDTO.getCar（）。getCarNumb（）））））&&（（（car.prevSchi不包含\“ N2 \” && car.prevSchi不包含\“ N1 \”）&&（car.prevLoadEmpty == \“ L \”）&&（（car.prevCarKind str [startsWith] \“ F \” || car.prevCarKind s tr [startsWith] \“ YF \”）&&（car.prevCarKind not str [startsWith] \“ FI \” && car.prevCarKind not str [startsWith] \“ FA \” && car.prevCarKind not str [startsWith] \“ FW \“ && car.prevCarKind not str [startsWith] \” FB \“）&&（car.prevCarKind！= \” YFB \“）））|| （（（car.nextSchi不包含\“ N2 \” && car.nextSchi不包含\“ N1 \”）&&（car.nextLoadEmpty == \“ L \”）&&（（（car.nextCarKind str [startsWith] \“ F \“ || car.nextCarKind str [startsWith] \” YF \“）&&（car.nextCarKind not str [startsWith] \” FA \“ && car.nextCarKind not str [startsWith] \” FB \“ && car.nextCarKind不是str [startsWith] \“ FI \” && car.nextCarKind不是str [startsWith] \“ FW \”）&&（car.nextCarKind！= \“ Y \” || car.nextCarKind！= \“ YFB \”））））

如果我们要将其拆分为不同的规则，以查找由于（prevCarKind，prevSchi，prevLoadEmpty）或（nextCarKind，nextSchi，nextLoadEmpty）是否满足该规则，将很难拆分复杂的规则。

我还给出了规则条件，就像从我们的UI应用程序中获取的图像一样。 Pictorial view of above rule condition string 如果有人可以让我们知道如何根据上述条件满足以上条件，这将是有帮助的。

关于， Madhankumar。 B

谢谢，玛丹

Answer 1

不，你不能。但是您可以将规则分成多个规则，如下所示：

rule "RUL1A"
when
    car.schi == 'A'
then
    ...
end

rule "RUL1B"
when
    car.carKind[0] == 'A'
then
   ...
end

Answer 2

您可以将条件分为几个功能（并在drl中定义它们）：

function boolean isKindEzzEngEtd(Car car) {
    return car.carKind in ["EZZ", "ENG", "ETD"];
}

function boolean isSchiContainsN1N2(Car car) {
    return (car.schi contains "N1" || car.schi contains "N2");
}

...

then use the functions inside your rules (it also improves readablility and re-use):

rule "RUL1"
when
   $car: Car() 
   !isKindEzzEngEtd($car) || isSchiContainsN1N2($car) && ...
then
    -- call the functions one by one to find out which one returned true:
    System.out.println("NOT isKindEzzEngEtd: " + !isKindEzzEngEtd($car));
    System.out.println("isSchiContainsN1N2: " + isSchiContainsN1N2($car));    
end

找出规则中所有条件元素均为真

2 个答案: