Dataweave JSON 到 XML format-Mule-4
Dataweave JSON to XML format-Mule-4
我是使用 dataweave 转换的初学者,我正在尝试编写一个 Mule 4 dataweave 表达式来将输入 JSON 转换为输出 XML 格式,我的有效负载为 JSON 格式,我想将其转换为特定的 XML 格式,下面是实际的 JSON 以及输出 XML
product-id(XML) 标签将来自 JSON 的 PBSI__Item__r 标签的 = name 属性
XML 中的分配值来自 PBSI__Inventory__r 标签的名称属性
JSON:
[
{
"PBSI__Item__r": {
"Id": null,
"type": "PBSI__PBSI_Item__c",
"Name": "116065"
},
"PBSI__Inventory__r": [
{
"Id": null,
"type": "PBSI__PBSI_Inventory__c",
"PBSI__Real_Quantity__c": "13.0"
}
],
"PBSI__Location__r": {
"Id": null,
"type": "PBSI__PBSI_Location__c",
"Name": "OB043"
},
"Id": null,
"type": "PBSI__Lot__c"
},
{
"PBSI__Item__r": {
"Id": null,
"type": "PBSI__PBSI_Item__c",
"Name": "116066"
},
"PBSI__Inventory__r": [
{
"Id": null,
"type": "PBSI__PBSI_Inventory__c",
"PBSI__Real_Quantity__c": "1.0"
}
],
"PBSI__Location__r": {
"Id": null,
"type": "PBSI__PBSI_Location__c",
"Name": "OA011"
},
"Id": null,
"type": "PBSI__Lot__c"
}
]
输出XML:
<?xml version='1.0' encoding='UTF-8'?>
<inventory xmlns="http://www.demandware.com/xml/impex/inventory/2007-05-31">
<inventory-list>
<header list-id="Hastens_Inventory">
<default-instock>false</default-instock>
<use-bundle-inventory-only>false</use-bundle-inventory-only>
</header>
<records>
<record product-id="116065">
<allocation>13</allocation>
<allocation-timestamp>2019-04-24T07:09:51.954Z</allocation-timestamp>
</record>
<record product-id="116066">
<allocation>1</allocation>
<allocation-timestamp>2019-04-24T07:09:51.965Z</allocation-timestamp>
</record>
</records>
</inventory-list>
</inventory>
这应该可以帮助您找到解决方案并学习解决方法,我还没有给出完整的解决方案,因此您可以以此为起点(已完成近 80%) .
除了包含命名空间外,请遵循 link https://docs.mulesoft.com/mule-runtime/4.3/dataweave-cookbook-include-xml-namespaces
{
inventory:{
inventorylist: {
header @("list-id":"Hastens_Inventory"):
{
"default-instock":false,
"use-bundle-inventory-only":false
},
records: {(payload map
{
record @("product-id": $.PBSI__Item__r.Name): {
allocation: $.PBSI__Inventory__r[0].PBSI__Real_Quantity__c,
"allocation-timestamp": now()
}
})}
}
}
}
我是使用 dataweave 转换的初学者,我正在尝试编写一个 Mule 4 dataweave 表达式来将输入 JSON 转换为输出 XML 格式,我的有效负载为 JSON 格式,我想将其转换为特定的 XML 格式,下面是实际的 JSON 以及输出 XML
product-id(XML) 标签将来自 JSON 的 PBSI__Item__r 标签的 = name 属性 XML 中的分配值来自 PBSI__Inventory__r 标签的名称属性
JSON:
[
{
"PBSI__Item__r": {
"Id": null,
"type": "PBSI__PBSI_Item__c",
"Name": "116065"
},
"PBSI__Inventory__r": [
{
"Id": null,
"type": "PBSI__PBSI_Inventory__c",
"PBSI__Real_Quantity__c": "13.0"
}
],
"PBSI__Location__r": {
"Id": null,
"type": "PBSI__PBSI_Location__c",
"Name": "OB043"
},
"Id": null,
"type": "PBSI__Lot__c"
},
{
"PBSI__Item__r": {
"Id": null,
"type": "PBSI__PBSI_Item__c",
"Name": "116066"
},
"PBSI__Inventory__r": [
{
"Id": null,
"type": "PBSI__PBSI_Inventory__c",
"PBSI__Real_Quantity__c": "1.0"
}
],
"PBSI__Location__r": {
"Id": null,
"type": "PBSI__PBSI_Location__c",
"Name": "OA011"
},
"Id": null,
"type": "PBSI__Lot__c"
}
]
输出XML:
<?xml version='1.0' encoding='UTF-8'?>
<inventory xmlns="http://www.demandware.com/xml/impex/inventory/2007-05-31">
<inventory-list>
<header list-id="Hastens_Inventory">
<default-instock>false</default-instock>
<use-bundle-inventory-only>false</use-bundle-inventory-only>
</header>
<records>
<record product-id="116065">
<allocation>13</allocation>
<allocation-timestamp>2019-04-24T07:09:51.954Z</allocation-timestamp>
</record>
<record product-id="116066">
<allocation>1</allocation>
<allocation-timestamp>2019-04-24T07:09:51.965Z</allocation-timestamp>
</record>
</records>
</inventory-list>
</inventory>
这应该可以帮助您找到解决方案并学习解决方法,我还没有给出完整的解决方案,因此您可以以此为起点(已完成近 80%) .
除了包含命名空间外,请遵循 link https://docs.mulesoft.com/mule-runtime/4.3/dataweave-cookbook-include-xml-namespaces
{
inventory:{
inventorylist: {
header @("list-id":"Hastens_Inventory"):
{
"default-instock":false,
"use-bundle-inventory-only":false
},
records: {(payload map
{
record @("product-id": $.PBSI__Item__r.Name): {
allocation: $.PBSI__Inventory__r[0].PBSI__Real_Quantity__c,
"allocation-timestamp": now()
}
})}
}
}
}