从 Map 获取列表并使用 groovy 进行过滤

Question

我正在尝试从地图中获取一个列表，并基于 id 的列表，并希望根据年龄（年龄 > 35）检索值（列表）

def people = [
"1": [[name:'Bob', age: 32, gender: 'M'],[name:'Johnny', age: 36, gender: 'M']],
"3": [[name:'Claire', age: 21, gender: 'F'],[name:'Amy', age: 54, gender:'F']],
"4": [[name:'John', age: 41, gender: 'F'],[name:'Sam', age: 54, gender:'F']]
]

def id = ["1","3"]


def age = people.subMap(id).values().findAll{it.age > 35}

但是，我收到以下错误：

Cannot compare java.util.ArrayList with value '[32, 36]' and java.lang.Integer with value '35'

如何获取所有年龄 > 35 岁的人员的名单？

Answer 1

您需要.flatten()的结果people.subMap(id).values()，因为它会生成映射列表的列表：

[[[name:Bob, age:32, gender:M], [name:Johnny, age:36, gender:M]], [[name:Claire, age:21, gender:F], [name:Amy, age:54, gender:F]]]

当您这样做时，people.subMap(id).values().flatten()您将获得地图列表：

[[name:Bob, age:32, gender:M], [name:Johnny, age:36, gender:M], [name:Claire, age:21, gender:F], [name:Amy, age:54, gender:F]]

然后你就可以申请了.findAll { it.age > 35 }

def people = [
        "1": [[name: 'Bob', age: 32, gender: 'M'], [name: 'Johnny', age: 36, gender: 'M']],
        "3": [[name: 'Claire', age: 21, gender: 'F'], [name: 'Amy', age: 54, gender: 'F']],
        "4": [[name: 'John', age: 41, gender: 'F'], [name: 'Sam', age: 54, gender: 'F']]
]

def id = ["1", "3"]

def age = people.subMap(id).values().flatten().findAll { it.age > 35 }

println age

输出：

[[name:Johnny, age:36, gender:M], [name:Amy, age:54, gender:F]]

希望能帮助到你。