Jackson XML序列化 – 移除字段标签

huangapple
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英文:

Jackson XML Serialization - Remove Field Tags

问题

I need to generate this XML:

  1. <CRequest>
  2.   <abc:Name>Smith</abc:Name>
  3.   <abc:FirstName>John</abc:FirstName>
  4.   <abc:Age>12</abc:Age>
  5.   <abc:Name>Jones</abc:Name>
  6.   <abc:FirstName>Jake</abc:FirstName>
  7.   <abc:Age>10</abc:Age>
  8.   <abc:Name>Johnson</abc:Name>
  9.   <abc:FirstName>Paul</abc:FirstName>
  10.   <abc:Age>12</abc:Age>
  11. </CRequest>

However, the best I could do was:

  1. <CRequest>
  2.   <children>
  3.     <abc:Name>Smith</abc:Name>
  4.     <abc:FirstName>John</abc:FirstName>
  5.     <abc:Age>12</abc:Age>
  6.   </children>
  7.   <children>
  8.     <abc:Name>Jones</abc:Name>
  9.     <abc:FirstName>Jake</abc:FirstName>
  10.     <abc:Age>10</abc:Age>
  11.   </children>
  12.   <children>
  13.     <abc:Name>Johnson</abc:Name>
  14.     <abc:FirstName>Paul</abc:FirstName>
  15.     <abc:Age>12</abc:Age>
  16.   </children>
  17. </CRequest>

I have the following Java classes:

  1. @JsonRootName("CRequest")
  2. @XmlAccessorType(XmlAccessType.FIELD)
  3. @JsonInclude(JsonInclude.Include.NON_EMPTY)
  4. public class ChildrenRequest {
  6.     @JacksonXmlElementWrapper(useWrapping = false)
  7.     private List<Child> children = new ArrayList<>();
  8.     ...

and

  1. @XmlAccessorType(XmlAccessType.FIELD)
  2. @JsonInclude(JsonInclude.Include.NON_EMPTY)
  3. @JsonPropertyOrder({"Name", "FirstName", "Age"})
  4. public class Child {
  6.     @JsonProperty("Name")
  7.     @JacksonXmlProperty(localName = "abc:Name")
  8.     private String name;
  10.     @JsonProperty("Surname")
  11.     @JacksonXmlProperty(localName = "abc:FirstName")
  12.     private String firstName;
  14.     @JsonProperty("Age")
  15.     @JacksonXmlProperty(localName = "abc:Age")
  16.     private String age;
  17.     ...

Is there a way to get rid of the children tags?

PS: Without "useWrapping = false" I get two children tags for every child.

英文:

I need to generate this XML:

  1. &lt;CRequest&gt;
  2.   &lt;abc:Name&gt;Smith&lt;/abc:Name&gt;
  3.   &lt;abc:FirstName&gt;John&lt;/abc:Surname&gt;
  4.   &lt;abc:Age&gt;12&lt;/abc:Age&gt;
  5.   &lt;abc:Name&gt;Jones&lt;/abc:Name&gt;
  6.   &lt;abc:FirstName&gt;Jake&lt;/abc:Surname&gt;
  7.   &lt;abc:Age&gt;10&lt;/abc:Age&gt;
  8.   &lt;abc:Name&gt;Johnson&lt;/abc:Name&gt;
  9.   &lt;abc:FirstName&gt;Paul&lt;/abc:Surname&gt;
  10.   &lt;abc:Age&gt;12&lt;/abc:Age&gt;
  11. &lt;/CRequest&gt;

However, the best I could do was:

  1. &lt;CRequest&gt;
  2.   &lt;children&gt;
  3.     &lt;abc:Name&gt;Smith&lt;/abc:Name&gt;
  4.     &lt;abc:FirstName&gt;John&lt;/abc:Surname&gt;
  5.     &lt;abc:Age&gt;12&lt;/abc:Age&gt;
  6.   &lt;/children&gt;
  7.   &lt;children&gt;
  8.     &lt;abc:Name&gt;Jones&lt;/abc:Name&gt;
  9.     &lt;abc:FirstName&gt;Jake&lt;/abc:Surname&gt;
  10.     &lt;abc:Age&gt;12&lt;/abc:Age&gt;
  11.   &lt;/children&gt;
  12.   &lt;children&gt;
  13.     &lt;abc:Name&gt;Johnson&lt;/abc:Name&gt;
  14.     &lt;abc:FirstName&gt;Paul&lt;/abc:Surname&gt;
  15.     &lt;abc:Age&gt;12&lt;/abc:Age&gt;
  16.   &lt;/children&gt;
  17. &lt;/CRequest&gt;

I have the following Java classes:

  1. @JsonRootName(&quot;CRequest&quot;)
  2. @XmlAccessorType(XmlAccessType.FIELD)
  3. @JsonInclude(JsonInclude.Include.NON_EMPTY)
  4. public class ChildrenRequest {
  6.     @JacksonXmlElementWrapper(useWrapping = false)
  7.     private List&lt;Child&gt; children= new ArrayList&lt;&gt;();
  8.     ...

and

  1. @XmlAccessorType(XmlAccessType.FIELD)
  2. @JsonInclude(JsonInclude.Include.NON_EMPTY)
  3. @JsonPropertyOrder({&quot;Name&quot;, &quot;FirstName&quot;, &quot;Age&quot;})
  4. public class Child{
  6.     @JsonProperty(&quot;Name&quot;)
  7.     @JacksonXmlProperty(localName = &quot;abc:Name&quot;)
  8.     private String name;
  10.     @JsonProperty(&quot;Surname&quot;)
  11.     @JacksonXmlProperty(localName = &quot;FirstName&quot;)
  12.     private String firstName;
  14.     @JsonProperty(&quot;Age&quot;)
  15.     @JacksonXmlProperty(localName = &quot;abc:Age&quot;)
  16.     private String age;
  17.     ...

Is there a way to get rid of the children tags?

PS: Without "useWrapping = false" I get two children tags for every child.

答案1

得分: 0

你需要实现请求类的自定义序列化器：

  1. class ChildrenRequestJsonSerializer extends JsonSerializer<ChildrenRequest> {
  2. 	@Override
  3. 	public void serialize(ChildrenRequest value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
  4. 		ToXmlGenerator xmlGen = (ToXmlGenerator) gen;
  5. 		writeStartObject(xmlGen);
  6. 		JsonSerializer<Object> childSerializer = serializers.findValueSerializer(Child.class).unwrappingSerializer(NameTransformer.NOP);
  7. 		for (Child child : value.getChildren()) {
  8. 			childSerializer.serialize(child, gen, serializers);
  9. 		}
  10. 		xmlGen.writeEndObject();
  11. 	}
  13. 	private void writeStartObject(ToXmlGenerator xmlGen) throws IOException {
  14. 		final XmlMapper mapper = (XmlMapper) xmlGen.getCodec();
  15. 		final PropertyName rootName = mapper.getSerializationConfig().findRootName(ChildrenRequest.class);
  17. 		xmlGen.setNextName(new QName("", rootName.getSimpleName()));
  18. 		xmlGen.writeStartObject();
  19. 	}
  20. }

然后，你可以像下面这样注册序列化器：

  1. @JsonRootName("CRequest")
  2. @JsonSerialize(using = ChildrenRequestJsonSerializer.class)
  3. class ChildrenRequest

请参考以下链接了解更多信息：

英文:

You need to implement custom serialiser for request class:

  1. class ChildrenRequestJsonSerializer extends JsonSerializer&lt;ChildrenRequest&gt; {
  2. 	@Override
  3. 	public void serialize(ChildrenRequest value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
  4. 		ToXmlGenerator xmlGen = (ToXmlGenerator) gen;
  5. 		writeStartObject(xmlGen);
  6. 		JsonSerializer&lt;Object&gt; childSerializer = serializers.findValueSerializer(Child.class).unwrappingSerializer(NameTransformer.NOP);
  7. 		for (Child child : value.getChildren()) {
  8. 			childSerializer.serialize(child, gen, serializers);
  9. 		}
  10. 		xmlGen.writeEndObject();
  11. 	}
  13. 	private void writeStartObject(ToXmlGenerator xmlGen) throws IOException {
  14. 		final XmlMapper mapper = (XmlMapper) xmlGen.getCodec();
  15. 		final PropertyName rootName = mapper.getSerializationConfig().findRootName(ChildrenRequest.class);
  17. 		xmlGen.setNextName(new QName(&quot;&quot;, rootName.getSimpleName()));
  18. 		xmlGen.writeStartObject();
  19. 	}
  20. }

And you can register serialiser as below:

  1. @JsonRootName(&quot;CRequest&quot;)
  2. @JsonSerialize(using = ChildrenRequestJsonSerializer.class)
  3. class ChildrenRequest

See also:

huangapple
  • 本文由 发表于 2020年8月4日 01:28:57
  • 转载请务必保留本文链接：https://go.coder-hub.com/63234236.html
  • jackson
  • java
  • serialization
  • xml
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