5

我正在尝试设置代码来创建一个节点树Jackson，然后可以使用它来编写JSON或XML. 我已经像这样手动创建了节点树：

XmlMapper nodeMapper = new XmlMapper();

ObjectNode rootNode = nodeMapper.createObjectNode();
ObjectNode currentNode = rootNode.putObject("Examples");
currentNode.put("Puppy", TRUE)
           .put("Apple", 2)
           .put("Jet", "Li");
currentNode = rootNode.putObject("Single");
currentNode.put("One", 1);

String writePath = "C:/users/itsameamario/Documents/basicXMLtest.xml";
nodeMapper.writeValue(new File(writePath), rootNode);

我的 XML 输出是：

<?xml version="1.0"?>
<ObjectNode>
    <Examples>
        <Puppy>true</Puppy>
        <Apple>2</Apple>
        <Jet>Li</Jet>
    </Examples>
    <Single>
        <One>1</One>
    </Single>
</ObjectNode>

但是对于 XML 的某些部分，我想向其中一个节点添加一个属性，如下所示：

<Examples overlyComplicated="yes">
<!--...-->
</Examples>

我发现的所有包含属性的示例都应用于预先存在的类。我一直无法找到将属性添加到上述手动构建的节点树的方法。使用它可行Jackson吗？

4

1 回答 1

8

attribute由于ObjectNode对序列化一无所知，因此无法标记给定属性。您可以对POJO类执行此操作，并且仅当注释用于给定属性com.fasterxml.jackson.dataformat.xml.ser.ToXmlGenerator时才会处理它。@JacksonXmlProperty(isAttribute = true)我建议为POJO需要属性的元素创建并使用Jackson XML注释或实现JsonSerializable接口。它可能如下所示：

import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializable;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.jsontype.TypeSerializer;
import com.fasterxml.jackson.databind.node.ObjectNode;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import com.fasterxml.jackson.dataformat.xml.ser.ToXmlGenerator;

import java.io.IOException;
import java.util.LinkedHashMap;
import java.util.Map;

public class XmlMapperApp {

    public static void main(String[] args) throws Exception {
        Map<String, Object> map = new LinkedHashMap<>();
        map.put("Puppy", Boolean.TRUE);
        map.put("Apple", 2);
        map.put("Jet", "Li");
        Examples examples = new Examples();
        examples.setOverlyComplicated("yes");
        examples.setMap(map);

        XmlMapper mapper = new XmlMapper();
        mapper.enable(SerializationFeature.INDENT_OUTPUT);

        ObjectNode rootNode = mapper.createObjectNode();
        rootNode.putPOJO("Examples", examples);
        ObjectNode currentNode = rootNode.putObject("Single");
        currentNode.put("One", 1);

        mapper.writeValue(System.out, rootNode);
    }
}

class Examples implements JsonSerializable {

    @Override
    public void serialize(JsonGenerator gen, SerializerProvider serializers) throws IOException {
        ToXmlGenerator toXmlGenerator = (ToXmlGenerator) gen;
        toXmlGenerator.writeStartObject();

        writeAttributes(toXmlGenerator);
        writeMap(toXmlGenerator);

        toXmlGenerator.writeEndObject();
    }

    private void writeAttributes(ToXmlGenerator gen) throws IOException {
        if (overlyComplicated != null) {
            gen.setNextIsAttribute(true);
            gen.writeFieldName("overlyComplicated");
            gen.writeString(overlyComplicated);
            gen.setNextIsAttribute(false);
        }
    }

    private void writeMap(ToXmlGenerator toXmlGenerator) throws IOException {
        for (Map.Entry<String, Object> entry : map.entrySet()) {
            toXmlGenerator.writeObjectField(entry.getKey(), entry.getValue());
        }
    }

    @Override
    public void serializeWithType(JsonGenerator gen, SerializerProvider serializers, TypeSerializer typeSer) throws IOException {
        serialize(gen, serializers);
    }

    private String overlyComplicated;
    private Map<String, Object> map;

    // getters, setters, toString
}

上面的代码打印：

<ObjectNode>
  <Examples overlyComplicated="yes">
    <Puppy>true</Puppy>
    <Apple>2</Apple>
    <Jet>Li</Jet>
  </Examples>
  <Single>
    <One>1</One>
  </Single>
</ObjectNode>

如果您想将其Example POJO用于JSON序列化，则需要在serialize方法中处理它或创建另一个ObjectNode而不是Examlples对象。

于 2019-07-17T00:02:31.627 回答