根据m子Dataweave中另一个列表中的值有效替换列表中的值

时间:2019-04-04 00:47:01

标签: arrays arraylist mule dataweave

需要一种有效的解决方案,用阵列2中的值替换阵列1(主列表)中的值。当前代码在Dataweave中工作。但是,当两个列表包含更多值时,我认为这不是最佳解决方案。建议使用从数组2中的值更新数组1的最佳解决方案。我正在使用Mule 3.8.5。我想要一个解决方案,以便只需要在第二个列表中进行一次迭代,并替换主列表中的所有值。当前,样本中两次使用了过滤条件。但是在实际情况下,我需要替换8个这样的值。

%dw 1.0
%output application/java
%var arr1 =[
    {
       "leadId": 127,
       "playerId": 334353,
       "name": "Joseph",
       "activity": "10/03/2015"
     },
     {
       "leadId": 128,
       "playerId": 334354,
       "name": "Sam",
       "activity": "10/03/2017"
     },
     {

      "leadId": 124,
       "playerId": "",
       "name": "",
        "activity": "10/03/2015"
     },
    {

      "leadId": 123,
       "playerId": "",
       "name": "",
        "activity": "10/03/2015"
     }

  ]
 %var arr2 =  [
   {
     "playerId": 123456,
     "name": "James",
     "leadId": 124
   },
   {
     "playerId": 7890,
     "name": "Jacob",
     "leadId": 123
   }
 ]

---
arr1 map ((actData) -> {
     "leadId":actData.leadId,
     "playerId": (arr2 filter ($.leadId == actData.leadId))[0].playerId default actData.playerId,
    "name": (arr2 filter ($.leadId == actData.leadId))[0].name default actData.name,
    "activity": actData.activity


})

上面的代码给出了预期的结果。但是当列表值更多时,它非常慢。

1 个答案:

答案 0 :(得分:1)

嗨,您应该使用groupBy,以便按条件将其编入索引,然后每次都进行查找而不是过滤器

%dw 1.0
%output application/json
%var arr1 =[
    {
       "leadId": 127,
       "playerId": 334353,
       "name": "Joseph",
       "activity": "10/03/2015"
     },
     {
       "leadId": 128,
       "playerId": 334354,
       "name": "Sam",
       "activity": "10/03/2017"
     },
     {

      "leadId": 124,
       "playerId": "",
       "name": "",
        "activity": "10/03/2015"
     },
    {

      "leadId": 123,
       "playerId": "",
       "name": "",
        "activity": "10/03/2015"
     }

  ]
 %var arr2 =  [
   {
     "playerId": 123456,
     "name": "James",
     "leadId": 124
   },
   {
     "playerId": 7890,
     "name": "Jacob",
     "leadId": 123
   }
 ]


%var arr2ByLeadId = arr2 groupBy $.leadId
---
arr1 map ((actData) -> {
    "leadId":actData.leadId,
    "playerId": arr2ByLeadId."$(actData.leadId)"[0].playerId default actData.playerId, 
    "name": arr2ByLeadId."$(actData.leadId)"[0].name default actData.name,
    "activity": actData.activity
})