开发XML转JSON-API的API服务，是否有适配的XSLT Schema？

Great question! Let's walk through how to tackle this, since converting simple XML to JSON-API via XSLT 3.0 is totally doable—even if there's no official "XSLT Schema" tailored specifically for this exact use case. Here's what you need to know:

JSON-API has strict structural rules (like top-level data, attributes, and relationships fields, standardized resource object formats, etc.). The key is to first map your input XML to an intermediate XML structure that aligns with JSON-API, then use XSLT 3.0's xml-to-json() function to convert that to valid JSON.

Step 1: Map Your XML to JSON-API's Structure #

Let's use a concrete example. Suppose your input XML looks like this:

<user>
  <id>123</id>
  <name>John Doe</name>
  <email>john@example.com</email>
  <postId>456</postId>
</user>

You'll write an XSLT stylesheet that transforms this into the structured XML expected by xml-to-json() (using the json namespace for type hints), while adhering to JSON-API specs:

<xsl:stylesheet version="3.0" 
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:json="http://www.w3.org/2005/xpath-functions/json">
  <xsl:output method="text"/>

  <xsl:template match="/">
    <!-- Build intermediate XML matching JSON-API structure -->
    <xsl:variable name="json-api-xml">
      <json:object>
        <data>
          <json:object>
            <type>
              <json:string>users</json:string>
            </type>
            <id>
              <json:string><xsl:value-of select="user/id"/></json:string>
            </id>
            <attributes>
              <json:object>
                <name>
                  <json:string><xsl:value-of select="user/name"/></json:string>
                </name>
                <email>
                  <json:string><xsl:value-of select="user/email"/></json:string>
                </email>
              </json:object>
            </attributes>
            <relationships>
              <json:object>
                <post>
                  <json:object>
                    <data>
                      <json:object>
                        <type>
                          <json:string>posts</json:string>
                        </type>
                        <id>
                          <json:string><xsl:value-of select="user/postId"/></json:string>
                        </id>
                      </json:object>
                    </data>
                  </json:object>
                </post>
              </json:object>
            </relationships>
          </json:object>
        </data>
      </json:object>
    </xsl:variable>

    <!-- Convert to formatted JSON -->
    <xsl:value-of select="xml-to-json($json-api-xml, map { 'indent': true() })"/>
  </xsl:template>
</xsl:stylesheet>

Step 2: Reuse Generic JSON-API Conversion Logic #

If you're handling multiple resource types (users, posts, comments, etc.), abstract reusable templates to avoid repetition:

• A template for generating resource objects (handles type, id, and attributes)
• A template for building relationships from nested or referenced XML elements
  This makes your stylesheet easier to maintain as your API grows.

Are There Pre-Built XSLT Solutions? #

While there's no official standardized XSLT schema or stylesheet for JSON-API, the community has shared open-source XSLT 3.0 snippets and templates that you can adapt:

• Generic templates that dynamically map XML elements to JSON-API attributes or relationships based on naming conventions
• Logic for handling collections (converting XML lists to JSON-API's data arrays)

Key Considerations #

• XSLT 3.0 Support: Make sure your processor (like Saxon HE/PE/EE, or Altova XMLSpy) fully supports XSLT 3.0 and the xml-to-json() function.
• JSON-API Details: Stick to JSON-API rules—like using plural type values, ensuring id is a string, and separating resource attributes from relationships.
• Nested XML: Don't flatten nested XML into attributes; map them to relationships instead to follow JSON-API's resource linkage pattern.

内容的提问来源于stack exchange，提问作者mkardakov