How to parse XML using Jackson API?

Question

I am using Jackson api to parse xml object.

<BESAPI xsi:noNamespaceSchemaLocation="BESAPI.xsd">
    <Employee Resource="https://abc:52311/api/employee/100"/>
    <Employee Resource="https://abc:52311/api/employee/200"/>
    <Employee Resource="https://abc:52311/api/employee/300"/>
    <Employee Resource="https://abc:52311/api/employee/400"/>
</BESAPI>

this is the structure of xml records. I want to get list of all resources as String. How can I achieve it using Jackson api?

By the way: Your XML example is invalid because it uses the xsi: prefix without defining it. You need to add xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" to the BESAPI root element. — Thomas Fritsch
– Thomas Fritsch, Commented May 29, 2018 at 20:17

Thomas Fritsch · Accepted Answer · 2018-05-29 22:10:12Z

First you need to write some Java classes modeling your XML content. The classes get @JacksonXml... annotations to tell Jackson the mapping between XML and Java. These annotations are especially important when a Java name is different from the XML name.

One class is for representing the <BESAPI> root element:

@JacksonXmlRootElement(localName = "BESAPI")
public class BESAPI {

    @JacksonXmlProperty(isAttribute = true, localName = "noNamespaceSchemaLocation", namespace = XMLConstants.W3C_XML_SCHEMA_INSTANCE_NS_URI)
    private String noNamespaceSchemaLocation;

    @JacksonXmlProperty(isAttribute = false, localName = "Employee")
    @JacksonXmlElementWrapper(useWrapping = false)
    private List<Employee> employees;

    // public getters and setters (omitted here for brevity)
}

and another class for representing the <Employee> element

public class Employee {

    @JacksonXmlProperty(isAttribute=true, localName="Resource")
    private String resource;

    // public getters and setters (omitted here for brevity)
}

Then you can use Jackson's XmlMapper for reading XML content.

XmlMapper xmlMapper = new XmlMapper();
File file = new File("example.xml");
BESAPI besApi = xmlMapper.readValue(file, BESAPI.class);
for (Employee employee : besApi.getEmployees()) {
    System.out.println(employee.getResource());
}

jurgen · Accepted Answer · 2018-06-23 13:32:55Z

0

SimpleXml can do this:

final String data = ...
final SimpleXml simple = new SimpleXml();
final Element element = simple.fromXml(data);
for (final Element employee : element.children) {
    System.out.println(employee.attributes.get("Resource"));
}

Will output:

https://abc:52311/api/employee/100
https://abc:52311/api/employee/200
https://abc:52311/api/employee/300
https://abc:52311/api/employee/400

From maven central:

<dependency>
    <groupId>com.github.codemonstur</groupId>
    <artifactId>simplexml</artifactId>
    <version>1.4.0</version>
</dependency>

answered Jun 23, 2018 at 13:32

jurgen

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