java - 从资源文件夹中读取文件

Question

我有一个 Maven 项目结构，其中有 src/main/resources/json/test.xml 文件我正在尝试使用以下代码读取它但无法读取它。我可能找不到指定的文件.我必须将文件对象传递给 unmarshal 函数，我如何使用其他方法执行此操作

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());

异常(exception)情况是

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

resources文件夹在构建时被maven合并到生成的classes文件夹中。因此，您可以通过以下方式获取该文件的 InputStream:

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

JAXB 可以从 InputStream 中解码，并且使用与定义类 YourClassName 的 ClassLoader 相同的规则定位文件. (有关详细信息，请参阅 getResourceAsStream。)