如何根据 groovy 中的条件从集合中删除 node/nodes

Question

如果同一日期有 "Action value exist 'U' and 'D' "，我将尝试删除所有节点。

它应该只留下下面示例列表中的前两个节点的列表，因为它没有 "D" 对 -Date:2015-03-16 、 Date:2015-03- 的操作17、Date:2015-03-22

已尝试 list.unique，但如果 U 和 D 都采取 exist.Please 建议的操作，则需要删除所有内容。

[
  [UserName:test, FileName:Santande.gpg, ServerPort:21, Success:true,Date:2015-03-16 06:28:46.12, Action:U],
 [UserName:test, FileName:Santande.gpg, ServerPort:21, Success:true, Date:2015-03-17 08:27:05.157, Action:U], 
 [UserName:test, FileName:Santande.gpg, ServerPort:21, Success:true, Date:2015-03-18 08:49:37.98, Action:U], 
 [UserName:test, FileName:Santande.gpg,  ServerPort:22, Success:true, Date:2015-03-18 14:00:42.23, Action:D],
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-18 15:00:42.321, Action:D], 
 [UserName:test, FileName:Santande.gpg,  ServerPort:21, Success:true, Date:2015-03-19 07:12:15.616, Action:U],
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-19 08:00:47.697, Action:D],
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-19 09:00:48.14, Action:D],
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-19 10:00:53.163, Action:D], 
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-20 09:00:47.373, Action:D], 
 [UserName:test, FileName:Santande.gpg, ServerPort:21, Success:true, Date:2015-03-20 09:08:51.637, Action:U], 
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-23 08:00:37.413, Action:D], 
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-23 09:00:46.423, Action:D],
 [UserName:test, FileName:Santande.gpg, ServerPort:21, Success:true, Date:2015-03-23 09:06:32.997, Action:U],
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-23 10:00:50.33, Action:D]
 [UserName:test, FileName:Santande.gpg, ServerPort:22, Success:true, Date:2015-03-22 10:00:50.33, Action:U]
]

Answer 1

按日期分组，然后选择不包含 U 和 D 条目的组。例如：

println([
 [UserName:'test', FileName:'Santande.gpg', ServerPort:'21', Success:'true', Date:'2015-03-16 06:28:46.120', Action:'U'],
 [UserName:'test', FileName:'Santande.gpg', ServerPort:'21', Success:'true', Date:'2015-03-17 08:27:05.157', Action:'U'], 
 [UserName:'test', FileName:'Santande.gpg', ServerPort:'21', Success:'true', Date:'2015-03-18 08:49:37.980', Action:'U'], 
 [UserName:'test', FileName:'Santande.gpg', ServerPort:'22', Success:'true', Date:'2015-03-18 14:00:42.230', Action:'D'],
].groupBy{ 
    it.Date.tokenize().first() 
}.findAll{ 
    it.value*.Action.toSet()!=['U','D'].toSet() 
}*.value.flatten())

如何根据 groovy 中的条件从集合中删除 node/nodes

How to remove node/nodes from a collection based on condition in groovy

groovy