Reading a file from resource folder

Question

i have a maven project structure where i have src/main/resources/json/test.xml file present i am trying to read this using the following code but not able to read it.I ma getting cannot find the file specified.I have to pass a file object to unmarshal function ,how can i do this using other apporach

File file = new File("json\\test.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(ServiceApi.class);

            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
            ServiceApi customer = (ServiceApi) jaxbUnmarshaller.unmarshal(file);
            System.out.println(customer.getService().size());

The exception is

javax.xml.bind.UnmarshalException
 - with linked exception:
[java.io.FileNotFoundException: C:\Users\jayesh_shah\Downloads\dbt-dataformstub\json\test.xml (The system cannot find the path specified)]
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl.java:202)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(UnmarshallerImpl.java:173)
    at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:137)
    at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:142)
    at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:151)
    at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:169)
    at com.ge.stub.api.jaxb.JAXBExample.main(JAXBExample.java:17)
Caused by: java.io.FileNotFoundException: C:\Users\jayesh_shah\Downloads\dbt-dataformstub\json\test.xml (The system cannot find the path specified)
    at java.io.FileInputStream.open(Native Method)
    at java.io.FileInputStream.<init>(FileInputStream.java:120)
    at java.io.FileInputStream.<init>(FileInputStream.java:79)
    at sun.net.www.protocol.file.FileURLConnection.connect(FileURLConnection.java:70)
    at sun.net.www.protocol.file.FileURLConnection.getInputStream(FileURLConnection.java:161)
    at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:653)
    at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:186)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:772)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:737)
    at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:119)
    at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1205)
    at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:522)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl.java:200)
    ... 6 more

Answer 1

The resources folder is merged into the generated classes folder by maven when building. So you can get an InputStream for that file via:

InputStream is = YourClassName.class.getResourceAsStream("/json/test.xml");

JAXB can unmarshal from InputStreams, and the file is located using the same rules as used by the ClassLoader which defined the class YourClassName. (See getResourceAsStream for more info.)