在Mule 4中合并JSON中的元素

Question

我有以下SQL Server选择查询：

SELECT s.RefId
          ,s.LocalId
          ,s.StateProvinceId
          ,s.SchoolName
          ,e.Email
          ,e.EmailType
      FROM SchoolInfo s
      LEFT OUTER JOIN SchoolEmail e
      ON    e.SchoolRefId = s.RefId
      WHERE s.RefId = :ref_id

在DataWeave中转换为JSON：

%dw 2.0
output application/json
---
payload

输出：

[
  {
    "StateProvinceId": "SA",
    "RefId": "7FDF722B-6BBA-4BF0-8205-A5380B269EF1",
    "EmailType": "prm",
    "LocalId": "1",
    "SchoolName": "Steve's School",
    "Email": "[email protected]"
  },
  {
    "StateProvinceId": "SA",
    "RefId": "7FDF722B-6BBA-4BF0-8205-A5380B269EF1",
    "EmailType": "sec",
    "LocalId": "1",
    "SchoolName": "Steve's School",
    "Email": "[email protected]"
  }
]

但是我希望将其与通用元素合并以生成所需的输出：

{
  "RefId": "7FDF722B-6BBA-4BF0-8205-A5380B269EF1",
  "LocalId": "1",
  "StateProvinceId": "SA",
  "SchoolName": "Steve's School",
  "Emails": [
    {
      "Email": "[email protected]",
      "EmailType": "prm"
    },
    {
      "Email": "[email protected]",
      "EmailType": "sec"
    }
  ]
}

我如何在M子4中做到这一点？

谢谢，史蒂夫

Answer 1

甚至是这个

%dw 2.0
output application/json
var Emails =  
    "Emails": payload map {
      Email: $.Email,
      EmailType: $.EmailType
}

---
payload distinctBy($.StateProvinceId) map {
  StateProvinceId: $.StateProvinceId,
  RefId: $.RefId,
  Emails:  Emails.Emails
} 

Answer 2

(payload  map {
  StateProvinceId: $.StateProvinceId,
  RefId: $.RefId,
  Emails:  Emails.Emails
})[0]

Answer 3

我希望这能给你这个主意。

%dw 2.0
output application/json
var Emails =  [
    "Emails": payload map {
      Email: $.Email,
      EmailType: $.EmailType
}
]
---
payload distinctBy($.StateProvinceId) map {
  StateProvinceId: $.StateProvinceId,
  RefId: $.RefId,
  Emails:  Emails.Emails[0]
}   

Answer 4

这可能会为您提供更通用的方法：

%dw 2.0
output application/json
fun returnEmail() =  
      payload groupBy($.StateProvinceId) mapObject {
         ($$): $  map {
               Email: $.Email,
               EmailType: $.EmailType
         }
} 


---
payload distinctBy($.StateProvinceId) map {
  StateProvinceId: $.StateProvinceId,
  RefId: $.RefId,
  Emails: returnEmail()[$.StateProvinceId]
}

Answer 5

我稍微修改了有效负载，在其中添加了另一个ProvinceId。

Answer 6

@@ Salim Khan的答案很正确。只需加reduce即可将数组转换为对象，作为与您的响应格式完全匹配的最后一位]

%dw 2.0
var email =   payload map (item,index) -> { 
"Email":item.Email,
"EmailType":item.EmailType }

output application/json
---
 payload distinctBy($.StateProvinceId) map  {
"StateProvinceId": $.StateProvinceId,
"RefId": $.RefId,
 "LocalId":  $.LocalId,
"SchoolName": $.SchoolName,
"Emails":email } reduce $