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Getting my JSP properties file - please hel  RSS feed

 
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Hi there!
Please spare some of your precious time to help me with this. I think it might be a basic Java problem but maybe a JSP/servlet problem too. Here is what I'm trying to do:
I have a jsp which runs off the Tomcat root context in my home machine as follows E:\Tomcat33\webapps\ROOT\mysite\myjsp.jsp
I then want to know the current directory where this jsp is running. I use application.getRealPath(request.getServletPath()) and I get it as: E:\Tomcat33\webapps\ROOT\mysite\myjsp.jsp
I then want to load a properties file using the above location which means that I want the path : E:\Tomcat33\webapps\ROOT\mysite\myprop.properties
I'm using this technique to avoid hardcoding.
How could I then manipulate: E:\Tomcat33\webapps\ROOT\mysite\myjsp.jsp into E:\Tomcat33\webapps\ROOT\mysite\myprop.properties? I have looked at the String class but none of the methods there seem to help me.
Your help would be greatly appreciated.
 
Greenhorn
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Use random access file for property file.Get the path and store it in string append that string to file.
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How could I then manipulate:
E:\Tomcat33\webapps\ROOT\mysite\myjsp.jsp into
E:\Tomcat33\webapps\ROOT\mysite\myprop.properties
1. Use the String method
int p = lastIndexOf( '\\' ) to find the last separator.
2. Add the file name
String pth = jspPath.substring(0,p ) + "\\myprop.properties" ;
That really should be File.separatorChar instead of '\\'
Bill

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ernest fakudze
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Oh, thank you very very much for your help guys! Now I can have good nights since this thing is done. Really appreciate.
 
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