Question

考虑一种方法

 def public Set<AgeRange> getAgeRanges(boolean excludeSenior) {
           -- something ---
     }

如何为此编写ExpandoMetaClass

    ClassName.metaClass.methodName << { boolean excludeSenior->
           -- something ---
       }

Answer 1

我在Groovy控制台中试过这个并且它有效：

class Something {

    Set getAgeRanges(boolean excludeSenior) {
        [1, 2, 3] as Set
    }
}

// test the original method
def s = new Something()
assert s.getAgeRanges(true) == [1, 2, 3] as Set

// replace the method
Something.metaClass.getAgeRanges = { boolean excludeSenior ->
    [4, 5, 6] as Set
}

// test the replacement method
s = new Something()
assert s.getAgeRanges(true) == [4, 5, 6] as Set

Answer 2

我不确定你在问什么，但这可能证明你在寻找什么：

某些课程：

class SomeClass {
    List<Integer> ages
}

为类添加方法的一些元编程：

SomeClass.metaClass.agesOlderThan { int minimumAge ->
    // note that "delegate" here will be the instance
    // of SomeClass which the agesOlderThan method
    // was invoked on...
    delegate.ages?.findAll { it > minimumAge }
}

创建SomeClass的实例并调用新方法...

def sc = new SomeClass(ages: [2, 12, 22, 32])

assert sc.agesOlderThan(20) == [22, 32]

这有帮助吗？

Grails：ExpandoMetaClass用于方法

2 个答案: