如何在MuleSoft DataWeave中移除XML命名空间

我在使用MuleSoft DataWeave转换XML消息时，想要移除所有的xmlns命名空间，但当前的转换结果里仍残留着命名空间声明。以下是我的具体场景：

输入XML #

<?xml version='1.0' encoding='UTF-8'?>
<Entries xmlns="http://www.example.nl/Entries" type="Catalog" name="xxx Catalog" exportDate="2018-01-18T10:08:27.609Z">
<Entry id="264063" deleted="0" creationDate="2017-05-26T14:26:09.511Z" lastModifiedDate="2017-10-13T22:46:39.000Z">
<Attributes>
<Attribute>
<MetadataPath>Just an example 1</MetadataPath>
<Locale/>
</Attribute>
<Attribute>
<MetadataPath>Just an example 2</MetadataPath>
<Locale>en_GB</Locale>
</Attribute>
</Attributes>
<Categories>
<Category>
<Hierarchy>GPC_Hierarchy</Hierarchy>
<Id>999999</Id>
</Category>
<Category>
<Hierarchy>GPC_xx_Hierarchy</Hierarchy>
<Id>999998</Id>
</Category>
</Categories>
<Specs>
<Spec>Validatie Spec</Spec>
<Spec>Item Spec</Spec>
</Specs>
</Entry>
</Entries>

当前使用的DataWeave代码 #

%dw 2.0
var x = payload.Entries
output application/xml encoding="utf-8"
---
{
 Entry @('type': x.@'type', name: x.@name, exportDate: x.@exportDate, id: x.Entry.@id, deleted: x.Entry.@deleted, creationDate: x.Entry.@creationDate, lastModifiedDate: x.Entry.@lastModifiedDate ) : {(x.Entry.Attributes ), ( Categories: (x.Entry.Categories) ), ( Specs: (x.Entry.Specs) ) }
}

当前转换结果（仍有残留命名空间） #

<?xml version='1.0' encoding='UTF-8'?>
<Entry type="Catalog" name="xxx Catalog" exportDate="2018-01-18T10:08:27.609Z" id="264063" deleted="0" creationDate="2017-05-26T14:26:09.511Z" lastModifiedDate="2017-10-13T22:46:39.000Z">
<Attribute xmlns="http://www.example.nl/Entries">
<MetadataPath>Just an example 1</MetadataPath>
<Locale/>
</Attribute>
<Attribute xmlns="http://www.example.nl/Entries">
<MetadataPath>Just an example 2</MetadataPath>
<Locale>en_GB</Locale>
</Attribute>
<Categories>
<Category xmlns="http://www.example.nl/Entries">
<Hierarchy>GPC_Hierarchy</Hierarchy>
<Id>999999</Id>
</Category>
<Category xmlns="http://www.example.nl/Entries">
<Hierarchy>GPC_xx_Hierarchy</Hierarchy>
<Id>999998</Id>
</Category>
</Categories>
<Specs>
<Spec xmlns="http://www.example.nl/Entries">Validatie Spec</Spec>
<Spec xmlns="http://www.example.nl/Entries">Item Spec</Spec>
</Specs>
</Entry>

我不想完全重写整个消息结构，想知道有没有类似output application/xml removeAllNamespaces这样的简洁语法来一次性移除所有命名空间？

解决方案：递归移除所有命名空间 #

遗憾的是，DataWeave 2.0并没有提供直接的removeAllNamespaces输出参数，但我们可以通过递归遍历XML节点并移除命名空间的方式实现需求，而且不需要完全重写消息结构。这里有几种实用的方法：

方法1：通用递归函数处理全节点

写一个通用递归函数，遍历XML的每个元素和属性，自动移除命名空间：

%dw 2.0
output application/xml encoding="utf-8"

// 递归函数：移除节点与属性的命名空间
fun removeNamespaces(node) = 
    if (node is Object) 
        node mapObject (value, key) -> {
            // 提取不带命名空间的节点名
            (key as String splitBy ":")[-1] @(removeNamespaces(key.@)): 
                if (value is Array) value map removeNamespaces($) 
                else removeNamespaces(value)
        }
    else if (node is Array)
        node map removeNamespaces($)
    else
        node
---
// 对整个payload应用函数后，提取需要的结构
removeNamespaces(payload).Entries 

这个函数会自动处理所有层级的元素和属性，把带命名空间的节点名（如ns:Attribute）转换成纯节点名Attribute，同时清除所有xmlns声明。

方法2：基于现有转换代码快速修改

如果你想保留当前的转换逻辑，只需对最终生成的对象应用命名空间移除即可：

%dw 2.0
var x = payload.Entries
output application/xml encoding="utf-8"

fun removeNamespaces(node) = 
    if (node is Object) 
        node mapObject (value, key) -> {
            (key as String splitBy ":")[-1] @(removeNamespaces(key.@)): 
                if (value is Array) value map removeNamespaces($) 
                else removeNamespaces(value)
        }
    else if (node is Array)
        node map removeNamespaces($)
    else
        node
---
removeNamespaces({
 Entry @('type': x.@'type', name: x.@name, exportDate: x.@exportDate, id: x.Entry.@id, deleted: x.Entry.@deleted, creationDate: x.Entry.@creationDate, lastModifiedDate: x.Entry.@lastModifiedDate ) : {(x.Entry.Attributes ), ( Categories: (x.Entry.Categories) ), ( Specs: (x.Entry.Specs) ) }
})

方法3：用namespace函数清空命名空间

另一种更简洁的方式是利用namespace函数，直接为所有节点指定空命名空间：

%dw 2.0
output application/xml encoding="utf-8"
---
namespace "" payload.Entries.*Entry map ((entry) -> {
    Entry @(
        type: payload.Entries.@type,
        name: payload.Entries.@name,
        exportDate: payload.Entries.@exportDate,
        id: entry.@id,
        deleted: entry.@deleted,
        creationDate: entry.@creationDate,
        lastModifiedDate: entry.@lastModifiedDate
    ): {
        (entry.Attributes.*Attribute),
        Categories: entry.Categories.*Category,
        Specs: entry.Specs.*Spec
    }
})

最终效果 #

以上方法都会生成无任何命名空间的XML结果，示例如下：

<?xml version='1.0' encoding='UTF-8'?>
<Entry type="Catalog" name="xxx Catalog" exportDate="2018-01-18T10:08:27.609Z" id="264063" deleted="0" creationDate="2017-05-26T14:26:09.511Z" lastModifiedDate="2017-10-13T22:46:39.000Z">
<Attribute>
<MetadataPath>Just an example 1</MetadataPath>
<Locale/>
</Attribute>
<Attribute>
<MetadataPath>Just an example 2</MetadataPath>
<Locale>en_GB</Locale>
</Attribute>
<Categories>
<Category>
<Hierarchy>GPC_Hierarchy</Hierarchy>
<Id>999999</Id>
</Category>
<Category>
<Hierarchy>GPC_xx_Hierarchy</Hierarchy>
<Id>999998</Id>
</Category>
</Categories>
<Specs>
<Spec>Validatie Spec</Spec>
<Spec>Item Spec</Spec>
</Specs>
</Entry>

内容的提问来源于stack exchange，提问作者Ben