根据条件和日期过滤元素答案

【问题标题】：Filtering elements based on condition and date filed根据条件和日期过滤元素
【发布时间】：2021-02-11 13:26:33
【问题描述】：

我正在尝试根据以下条件过滤我的列表。

检查是否有重复的name 和employee 字段的记录。如果有重复，则只保留 date 字段值为最新的元素，忽略其他重复值。

输入列表：

[{
    name: "Bob",
    employee: "yes",
    date: "10/12/2020"
},
{
    name: "John",
    employee: "no",
    date: "14/12/2020"
},
{
    name: "Bob",
    employee: "yes",
    date: "14/12/2020"
},
{
    name: "Doe",
    employess: "yes",
    date: "10/01/2020"
},
{
    name: "Bob",
    employess: "yes",
    date: "09/12/2020"
},
{
    name: "Doe",
    employee: "yes",
    date: "01/12/2019"
},
{
    name: "Bob",
    employee: "no",
    date: "01/12/2019"
}]

理想的输出：

[{
    name: "Bob",
    employee: "yes",
    date: "14/12/2020"
},
{
    name: "John",
    employee: "no",
    date: "14/12/2020"
},
{
    name: "Bob",
    employee: "no",
    date: "01/12/2019"
},
{
    name: "Doe",
    employess: "yes",
    date: "10/01/2020"
}]

我尝试了findAll() 方法，但无法获得想要的结果。

【问题讨论】：

显示您的尝试以及失败的原因，以便我们为您提供帮助。

标签： java grails groovy filter

【解决方案1】：

你可以groupByname和employee的子图然后找到 max date 在分组项的值中。例如

def data = [
    [a: 1, b: 2, c: 0],
    [a: 1, b: 2, c: 3],
    [a: 2, b: 3, c: 0],
]

data.groupBy{
    it.subMap(["a", "b"])
}.collect{ k, v -> 
    k + [c: v*.c.max()] 
}
// -> [[a:1, b:2, c:3], [a:2, b:3, c:0]]

【讨论】：

【解决方案2】：

一个简单的inject 会有所帮助：

def list = [[
    name: "Bob",
    employee: "yes",
    date: "10/12/2020"
],
[
    name: "John",
    employee: "no",
    date: "14/12/2020"
],
[
    name: "Bob",
    employee: "yes",
    date: "14/12/2020"
],
[
    name: "Doe",
    employee: "yes",
    date: "10/01/2020"
],
[
    name: "Bob",
    employee: "yes",
    date: "09/12/2020"
],
[
    name: "Doe",
    employee: "yes",
    date: "01/12/2019"
],
[
    name: "Bob",
    employee: "no",
    date: "01/12/2019"
]]

def sdf = new java.text.SimpleDateFormat('dd/MM/yyyy')

def result = list.inject( [:] ){ res, curr ->
  Date date = sdf.parse curr.date
  String key = curr.name + '_' + curr.employee
  if( res[ key ]?.date < date ){
    res[ key ] = curr + [ date:date ]
  }
  res
}.values().collect{ it + [ date:sdf.format( it.date ) ] }.join('\n')

println result

打印

[name:Bob, employee:yes, date:14/12/2020]
[name:John, employee:no, date:14/12/2020]
[name:Doe, employee:yes, date:10/01/2020]
[name:Bob, employee:no, date:01/12/2019]

【讨论】：

【解决方案3】：

作为一个选项，您可以将初始列表收集为嵌套地图。例如，首先您可以按名称对记录进行分组。然后按员工分组。最终，您将获得以下 Map。现在您可以通过地图中的键和列表中的最新日期来获取记录。

【讨论】：