所以,我正在尝试使用docBuilder.parse(filepath)
用Java打开XML文档>
这是我的代码的样子:
public void openXMLfile(String filepath) { try { DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance(); DocumentBuilder docBuilder = docFactory.newDocumentBuilder(); Document doc = docBuilder.parse(filepath); } catch(Exception e) { e.printStackTrace(); } }
我已写出文件的绝对路径为:
[C:\\Users\\"my User account"\\Desktop
,是“我的用户帐户”,由我的用户帐户的实际名称代替
在我的主要功能中,看起来像这样:
public static void main(String[] args) { App aplication = new App(); String filepath = "C:\\Users\\"my User account"\\Desktop"; aplicacao.openXMLfile(filepath); }
但是,出现以下错误:
java.net.MalformedURLException: unknown protocol: c at java.net.URL.<init>(Unknown Source) at java.net.URL.<init>(Unknown Source) at java.net.URL.<init>(Unknown Source) at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(Unknown Source) at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(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.DOMParser.parse(Unknown Source) at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(Unknown Source) at javax.xml.parsers.DocumentBuilder.parse(Unknown Source) at Trabalho.App.openXMLfile(App.java:39) at Trabalho.App.main(App.java:70)
有人能说明我做错了什么吗?可能是文件路径本身吗?
非常感谢!
因此,我正尝试使用docBuilder.parse(filepath)使用Java打开XML文档,这就是我的代码:public void openXMLfile(String filepath){试试{...
此问题给出示例-How should a file: URI corresponding to a Windows path name look like?