如何解决 [org.xml.sax.SAXParseException;行号：1；列号：1；序言中不能有内容。]

Question

以下是我的 xml 示例 enter link description here

我的编码是

JAXBContext jaxbContext = JAXBContext.newInstance(NewsMLObj.class);
        SAXParserFactory spf = SAXParserFactory.newInstance();
        XMLReader xr = spf.newSAXParser().getXMLReader();

        // to bypass XML DocType and Entity as Jap did not provide proper XML
        xr.setFeature("http://xml.org/sax/features/validation", false);
        xr.setFeature("http://apache.org/xml/features/nonvalidating/load-dtd-grammar", false);
        xr.setFeature("http://apache.org/xml/features/nonvalidating/load-external-dtd", false);
        xr.setFeature("http://xml.org/sax/features/external-general-entities", false);
        xr.setFeature("http://xml.org/sax/features/external-parameter-entities", false);
        xr.setFeature("http://xml.org/sax/features/use-entity-resolver2", false);

        InputSource is = new InputSource(new FileReader(factoryType.serverXML.getInputFile2() + filename));
        SAXSource source = new SAXSource(xr, is);
        out.println("input source=" + is);
        javax.xml.bind.Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
        out.println("jaxbUnmarshaller =" + jaxbUnmarshaller);
        NewsMLObj nmo = (NewsMLObj) jaxbUnmarshaller.unmarshal(source);

运行“nmo”时，出现错误“javax.xml.bind.UnmarshalException - 有关联的例外： [org.xml.sax.SAXParseException;行号：1；列号：1；序言中不允许有内容。]"

javax.xml.bind.UnmarshalException
- with linked exception:
[org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 1; Content is not allowed in prolog.]
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.createUnmarshalException(Unknown Source)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.createUnmarshalException(Unknown Source)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(Unknown Source)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(Unknown Source)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(Unknown Source)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(Unknown Source)
at com.n2n.NekkeiFlashNews.client.imp.PacketToObjectNewsHostServer.processRawNews(PacketToObjectNewsHostServer.java:83)
at com.n2n.NekkeiFlashNews.client.imp.NewsRawFileReceiverThread.run(NewsRawFileReceiverThread.java:57)
at java.lang.Thread.run(Unknown Source)
Caused by: org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 1;   Content is not allowed in prolog.
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(Unknown Source)
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.fatalError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLScanner.reportFatalError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl$PrologDriver.next(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(Unknown Source)
... 7 more

是我的编码有问题吗？如何解决我的问题？

感谢和最好的问候 沙龙

Answer 1

错误消息“Prolog 中不允许内容”的出现可能有多种原因。这基本上意味着解析器在成功读取文档中的第一个有意义的内容之前发现了错误。这可能是（正如消息所暗示的那样）因为文档以“

我会先检查一下

new FileReader(factoryType.serverXML.getInputFile2() + filename)

返回可用于读取内容的 Reader，无需将该内容提交给 XML 解析。

Answer 2

看来，您的 xml 文件在序言之前写入了一些数据。 字符串之前应该没有任何内容，看起来像这样：

<?xml version="1.0" encoding="UTF-8"?>