用于评估 Java 对象的逻辑表达式的库

Ale*_*lex 4 java expression evaluate

假设我有以下课程:

class Person {
    int age;
    String city;
    Collection<Person> friends;
    Person spouse;
}
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我需要一个库,它允许我评估给定 Person 对象上的逻辑表达式是否为真。表达式看起来像这样:

((age>25 OR spouse.age>27) AND city=="New-York" AND size(friends)>100)
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所以,要求是:

  1. 能够使用基本逻辑运算符
  2. 访问给定对象的属性
  3. 访问内部对象的属性
  4. 使用简单的数学\类似 SQL 的函数,例如 size、max、sum

建议?

ass*_*ias 5

您可以使用ScriptEngine + 反射:

这是一个输出的人为示例:

age = 35
city = "London"
age > 32 && city == "London" => true
age > 32 && city == "Paris" => false
age < 32 && city == "London" => false
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如果您想处理非原始类型(例如集合),它可能会变得非常混乱。

public class Test1 {

    public static void main(String[] args) throws Exception{
        Person p = new Person();
        p.age = 35;
        p.city = "London";

        ScriptEngineManager factory = new ScriptEngineManager();
        ScriptEngine engine = factory.getEngineByName("JavaScript");

        Class<Person> c = Person.class;
        for (Field f : c.getDeclaredFields()) {
            Object o = f.get(p);
            String assignement = null;
            if (o instanceof String) {
                assignement = f.getName() + " = \"" + String.valueOf(o) + "\"";
            } else {
                assignement = f.getName() + " = " + String.valueOf(o);
            }
            engine.eval(assignement);
            System.out.println(assignement);
        }

        String condition = "age > 32 && city == \"London\"";
        System.out.println(condition + " => " + engine.eval(condition));

        condition = "age > 32 && city == \"Paris\"";
        System.out.println(condition + " => " + engine.eval(condition));

        condition = "age < 32 && city == \"London\"";
        System.out.println(condition + " => " + engine.eval(condition));
    }

    public static class Person {

        int age;
        String city;
    }
}
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