问题描述
我有一个位于 D:\XML\RequestXML
中的xml文件,我正在从 FileReader
。在我的程序中,我硬编码了文件路径 / XML / RequestXML /
。这对 windows
环境可以正常工作。在windows JBoss
中 D:\ jbossdistrib \jboss
。
我在 linux
/ usr / XML / RequestXML /
中创建了文件夹结构。并将xml添加到RequestXML文件夹中。 JBoss
位于 / usr / jbossdistrib / jboss /
路径中。
如果我将文件路径更改为,那么我们的应用程序就无法在/ XML / RequestXML / / usr / XML / RequestXML /
它可以在linux中运行。
如何在Linux和Windows中使用一致的文件路径?公共类控制器扩展HttpServlet {
私有字符串filePath =/ XML / RequestXML /; $ p
$ b保护void doPost(HttpServletRequest请求,
HttpServletResponse响应)抛出ServletException,IOException {
String file = request.getParameter(fileName);
xml = readFile(filePath + file);
$ b private String readFile(String file){
StringBuffer fileData = new StringBuffer();
try {
BufferedReader reader = new BufferedReader(new FileReader(file));
char [] buf = new char [1024];
int numRead = 0; ((numRead = reader.read(buf))!= -1){
String readData = String.valueOf(buf,0,numRead);
$ b $
fileData.append(readData);
buf = new char [1024];
}
reader.close();
$ b catch(FileNotFoundException e){
logger.fatal(在指定路径中找不到文件+文件);
catch(IOException e){
logger.fatal(读取xml文件时出错);
}
return fileData.toString();
更新
我的问题是如何在没有 / usr /
的情况下设置文件路径,这在Windows中可以正常工作。
如果这是不可能的,那么我是否需要在Windows环境中使用路径 / usr / XML / RequestXML /
?所以我必须在Windows中创建一个像 D:\usr\XML\RequestXML
的文件夹结构。
解决方案如果您知道当前的工作目录(测试它:
System.out .println(new File(。)。getAbsolutePath());
目录像../../ XML / RequestXML
记录:虽然这可能有帮助,但我仍然相信你应该试着用配置参数来解决这个问题,加载它作为类路径中的一个资源。
I have a xml file located in D:\XML\RequestXML
and I am reading xml file in this folder from a FileReader
. In my program I hard coded the file path /XML/RequestXML/
. This works fine with the windows
environment. In windows JBoss
is in D:\jbossdistrib\jboss
.
I created the folder structure in linux
/usr/XML/RequestXML/
. And add the xml in to RequestXML folder. JBoss
is in /usr/jbossdistrib/jboss/
path.
But my application can not find the file specified in /XML/RequestXML/ in linux environment.
If I change the file path as /usr/XML/RequestXML/
it works in linux.
How can I use the consistent file path in linux and windows both?
public class Controller extends HttpServlet {
private String filePath = "/XML/RequestXML/";
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
String file = request.getParameter("fileName");
xml = readFile(filePath + file);
}
private String readFile(String file) {
StringBuffer fileData = new StringBuffer();
try {
BufferedReader reader = new BufferedReader(new FileReader(file));
char[] buf = new char[1024];
int numRead=0;
while((numRead=reader.read(buf)) != -1){
String readData = String.valueOf(buf, 0, numRead);
fileData.append(readData);
buf = new char[1024];
}
reader.close();
}
catch (FileNotFoundException e) {
logger.fatal("File not found in specifid path "+ file);
}
catch (IOException e) {
logger.fatal("Error while reading the xml file");
}
return fileData.toString();
}
}
Update
My question is how to set the file path without /usr/
which works fine in Windows.If this is not possible, then do I need to use the path as /usr/XML/RequestXML/
in windows environment as well? so I have to create a folder structure like D:\usr\XML\RequestXML
in windows.
解决方案 If you know the current working directory (test it with:
System.out.println(new File(".").getAbsolutePath());
you can hardcode a relative directory like ../../XML/RequestXML
For the record: although this may help, I still believe you should try to solve this with a configuration parameter or by loading it as a resource available in the classpath.
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