将多个JSON对象转换为一个列表Mule dataweave

Question

我有如下输入

 { "ResidentialAddress": { "type": "RES", "Building": "String" }, "Name": "Satheesh", "Status": "Active", "OfficeAddress": { "type": "OFC", "Building": "String" }, "TempAddress": { "type": "TEMP", "Building": "String" } } 

我希望将其转换如下

 {Address:[ { "type": "RES", "Building": "String" }, { "type": "OFC", "Building": "String" },{ "type": "TEMP", "Building": "String" } ]} 

当我尝试使用address：payload.ResidentialAddress ++ payload.TempAddress时，它给了我组合的字段，而不是列表，谁能帮忙？

Answer 1

在dataweave中使用扁平化操作。 在dataweave设置有效负载之后。

     %dw 1.0
 %output application/json
 ---
(flatten payload) filter ($.type != null)

最终的xml文件将是

<dw:transform-message doc:name="Transform Message">
            <dw:set-payload><![CDATA[ %dw 1.0
 %output application/json
 ---
 (flatten payload) filter ($.type != null)]]></dw:set-payload>
        </dw:transform-message>
        <set-payload value="{&quot;Address&quot;: #[payload]}" mimeType="application/json" doc:name="Set Payload"/>
        <json:object-to-json-transformer doc:name="Object to JSON"/>

输出：

Answer 2

这应该适用于过滤具有字符串/整数/布尔值的任何字段：

%dw 1.0
%output application/json
---

{
 Address: (flatten payload) filter ($ is :object)
}

Answer 3

针对您的测试数据：

%dw 1.0
%output application/json
---
{ Address:payload filter $ is :object map $}

给出：

 {
  "Address": [
    {
      "type": "RES",
      "Building": "String"
    },
    {
      "type": "OFC",
      "Building": "String"
    },
    {
      "type": "TEMP",
      "Building": "String"
    }
  ]
}

但您可能需要调整过滤条件才能使用真实数据...