CODEWITHSUNDEEP
javajacksonjackson-databindjackson-dataformat-xml
Eranda
Eranda

Reputation: 1467

Deserialize XML element with attributes with no value using Jackson Java

I am trying to deserialize the following XML and I couldn't get the parameter param section deserialized.

<video src="https://google.com/sample.mp4">
    <param>s</param>
    <param>Y</param>
    <param>Z</param>
</video>

My model

import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlProperty;

import java.util.ArrayList;
import java.util.List;

public class Video {
    @JacksonXmlProperty(isAttribute = true)
    private String src;

    @JacksonXmlElementWrapper(localName = "param", useWrapping = false)
    private List<String> param = new ArrayList<>();

    public String getSrc() {
        return src;
    }

    public List<String> getParam() {
        return param;
    }

    public void setParam(List<String> param) {
        this.param = param;
    }
}

Output

{
    "src": "https://google.com/sample.mp4",
    "param": [
        "Z"
    ]
}

I am expecting the values of param to be something like

{
    "src": "https://google.com/sample.mp4",
    "param": [
        "s",
        "Y",
        "Z"
    ]
}

Java code

ObjectMapper mapper = new ObjectMapper(new XmlFactory());
mapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY);
Video video = mapper.readValue(s, Video.class);
System.out.println(new ObjectMapper(new JsonFactory()).writeValueAsString(video));

Can someone help me getting it working. Thank you.

Upvotes: 1

Views: 2421

Answers (2)

Dmitry Pisklov
Dmitry Pisklov

Reputation: 1206

You need to use:

@JacksonXmlProperty(localName = "param")
@JacksonXmlElementWrapper(useWrapping = false)
private List<String> param = new ArrayList<>();

and delete mapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY); as it only masks the issue.

This code worked for me:

XmlMapper mapper = new XmlMapper();
Video video = mapper.readValue(s, Video.class);
System.out.println(new ObjectMapper(new JsonFactory()).writeValueAsString(video));

Outputs: {"src":"https://google.com/sample.mp4","param":["s","Y","Z"]}

Upvotes: 2

Nithin
Nithin

Reputation: 703

I used the following code and it worked for me,

XmlMapper mapper = new XmlMapper();
    mapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY);
    Video video = mapper.readValue(s, Video.class);
    System.out.println(new ObjectMapper(new JsonFactory()).writeValueAsString(video));

XmlMapper is frorm package com.fasterxml.jackson.dataformat.xml.XmlMapper

Hope it helped you.

Upvotes: 1

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