Question

File schemaLocation = new File("/home/localpc/servletaddequi/src/main/resources/BackupSample/DEMO.xsd")

如果我正在传递如上所述的文件路径，我可以获取文件。但我只想从项目传递文件的路径，我怎么能通过文件的路径？如果我传递文件路径如下：

File schemaLocation = new File("/foldername/filename.xsd");

它会产生错误：

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

Answer 1

请参阅此帖子的回答：mule read single file from classpath during flow，您可以使用var map = { ... 'sax': 'node_modules/sax/lib' }; var packages = { ... 'sax': { main: 'sax.js', defaultExtension: 'js' } };获取Thread.currentThread().getContextClassLoader().getResource("filename.xsd")课程。

因此，您可以尝试以下代码：

URL

使用mule esb中的servlet从项目中获取java转换器中的文件路径

1 个答案: