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

Question

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

如果我按照上面提到的方式传递文件路径，我就可以获取文件。但是我只想传递项目中的文件路径，我该如何传递文件路径？如果我按如下方式传递文件路径：

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

它生成错误为：

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

参考这个post的回答：mule read single file from classpath during flow，你可以使用Thread.currentThread().getContextClassLoader().getResource("filename.xsd")得到URLclass.

因此你可以试试下面的代码：

URL url = Thread.currentThread().getContextClassLoader().getResource("filename.xsd");
File schemaLocation = new File(url.getFile());

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

Getting file path in java transformer from project using servlet in mule esb

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