如何使用DataWeave 2.0将JSON对象合并至嵌套JSON的指定节点

你需要的DataWeave 2.0 代码如下，默认Input1为当前请求payload，Input2存储在变量vars.input2中，可根据你实际的变量命名调整：

%dw 2.0
output application/json
---
payload map (item) -> {
    (item - "Org_Response"),
    Org_Response: {
        (item.Org_Response - "Response_Data"),
        Response_Data: item.Org_Response.Response_Data ++ vars.input2
    }
}

实现逻辑 #

• 遍历Input1外层数组的所有元素，保留除Org_Response外的所有原有字段
• 重构Org_Response节点，保留除Response_Data外的所有原有字段
• 使用DataWeave的++合并运算符，将Input2的全部内容直接合并到Response_Data节点下，不会修改其他层级的结构

输入输出参考 #

Input 1 #

[{"Org_Response": {"Request_Criteria": {"Org_Type_Reference": {"ID": {"type": "Org_Type","text": "Business_Unit"}},"Include_Inactive": "0"},"Response_Data": {"Org": {"Reference": {"ID": {"type": "Business_Unit_Reference_ID","text": "999-99-FD"}},"Org_Data": {"Reference_ID": "999-99-FD","Name": "Management"}}}}]

Input 2 #

{"Org": {"Reference": {"ID": {"type": "Business_Unit_Reference_ID","text": "90000"}},"Org_Data": {"Reference_ID": "90000","Name": "TAXES"}}}

输出结果 #

[{"Org_Response": {"Request_Criteria": {"Org_Type_Reference": {"ID": {"type": "Org_Type","text": "Business_Unit"}},"Include_Inactive": "0"},"Response_Data": {"Org": {"Reference": {"ID": {"type": "Business_Unit_Reference_ID","text": "999-99-FD"}},"Org_Data": {"Reference_ID": "999-99-FD","Name": "Management"},"Org": {"Reference": {"ID": {"type": "Business_Unit_Reference_ID","text": "90000"}},"Org_Data": {"Reference_ID": "90000","Name": "TAXES"}}}}]

内容的提问来源于stack exchange，提问作者user12277274