File schemaLocation = new File("/home/localpc/servletaddequi/src/main/resources/BackupSample/DEMO.xsd")
如果我像上面提到的那样传递文件路径，我就能得到文件。但是我只想从项目中传递文件的路径，我怎么能传递文件的路径?如果我传递文件路径如下:
File schemaLocation = new File("/foldername/filename.xsd");
生成如下错误:
Failed to read schema document 'file:/BackupSample/DEMO.xsd', because 1) could not find the document; 2) the document could not be read; 3) the root element of the document is not <xsd:schema>.
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(ErrorHandlerWrapper.java:198)
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(ErrorHandlerWrapper.java:134)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(XMLErrorReporter.java:437)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(XMLErrorReporter.java:347)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.reportSchemaErr(XSDHandler.java:4166)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.reportSchemaError(XSDHandler.java:4149)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.getSchemaDocument1(XSDHandler.java:2479)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.getSchemaDocument(XSDHandler.java:2187)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.parseSchema(XSDHandler.java:573)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadSchema(XMLSchemaLoader.java:617)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadGrammar(XMLSchemaLoader.java:575)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadGrammar(XMLSchemaLoader.java:541)
at com.sun.org.apache.xerces.internal.jaxp.validation.XMLSchemaFactory.newSchema(XMLSchemaFactory.java:252)
at javax.xml.validation.SchemaFactory.newSchema(SchemaFactory.java:627)
at javax.xml.validation.SchemaFactory.newSchema(SchemaFactory.java:643)
at com.test.Test.transformMessage(Test.java:38)
at org.mule.transformer.AbstractMessageTransformer.transform(AbstractMessageTransformer.java:141)
at org.mule.transformer.AbstractMessageTransformer.transform(AbstractMessageTransformer.java:89)
at org.mule.DefaultMuleMessage.transformMessage(DefaultMuleMessage.java:1602)
at org.mule.DefaultMuleMessage.applyAllTransformers(DefaultMuleMessage.java:1509)
和
Caused by: java.io.FileNotFoundException: /BackupSample/DEMO.xsd (No such file or directory)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:146)
at java.io.FileInputStream.<init>(FileInputStream.java:101)
at sun.net.www.protocol.file.FileURLConnection.connect(FileURLConnection.java:90)
at sun.net.www.protocol.file.FileURLConnection.getInputStream(FileURLConnection.java:188)
at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:619)
at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:189)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaParsingConfig.parse(SchemaParsingConfig.java:582)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaParsingConfig.parse(SchemaParsingConfig.java:685)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaDOMParser.parse(SchemaDOMParser.java:530)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.getSchemaDocument(XSDHandler.java:2175)
... 218 more
参考这篇文章的回答:mule在流程中从类路径读取单个文件，您可以使用Thread.currentThread().getContextClassLoader().getResource("filename.xsd")来获取URL类。
所以你可以试试下面的代码:
URL url = Thread.currentThread().getContextClassLoader().getResource("filename.xsd");
File schemaLocation = new File(url.getFile());