从资源文件夹中读取文件

Question

我有一个maven项目结构,我有src/main/resources/json/test.xml文件存在我试图使用以下代码读取它但无法读取它.我得到找不到指定的文件.我有要将文件对象传递给unmarshal函数,我该怎么做才能使用其他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());

例外是

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

在构建时,maven将资源文件夹合并到生成的类文件夹中.所以你可以InputStream通过以下方式获得该文件:

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

JAXB可以从InputStreams 解组,并且使用与ClassLoader定义类的规则相同的规则来定位文件YourClassName.(有关详细信息,请参阅getResourceAsStream.)