i tried this with tomcat 4.1.8 to 4.1.10 with JspServlet's config: <init-param> <param-name>compiler</param-name> <param-value>jikes</param-value> </init-param> getting this error: org.apache.jasper.JasperException: Unable to compile class for JSP
An error occurred at line: -1 in the jsp file: null Generated servlet error: [javac] Compiling 1 source file at org.apache.jasper.compiler.DefaultErrorHandler.javacError(DefaultErrorHandler.java:120) at org.apache.jasper.compiler.ErrorDispatcher.javacError(ErrorDispatcher.java:293) at org.apache.jasper.compiler.Compiler.generateClass(Compiler.java:315) at org.apache.jasper.compiler.Compiler.compile(Compiler.java:326) at org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:474) at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:182) at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:289) at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:240) at javax.servlet.http.HttpServlet.service(HttpServlet.java:853) ... in the log i got: 2002-09-01 20:13:09 Compiler jikes 2002-09-01 20:13:10 Error compiling file: C:\java\jakarta\tomcat\jakarta-tomcat-4.1.10\work\Standalone\localhost\_\/index_jsp.java [javac] Compiling 1 source file looking into the code i have found the method getServletJavaFileName() in JspCompilationContext: public String getServletJavaFileName() { if (servletJavaFileName != null) { return servletJavaFileName; } String outputDir = getOutputDir(); servletJavaFileName = getServletClassName() + ".java"; if (outputDir != null && !outputDir.equals("")) { if( outputDir.endsWith("/" ) ) { servletJavaFileName = outputDir + servletJavaFileName; } else { servletJavaFileName = outputDir + "/" + servletJavaFileName; } } return servletJavaFileName; } ... i guess since outputDir is only checked for "/" and not window's file separator "\" the path to the generated jsp becomes wrong. since i am not really a developer and more a tomcat user i would like to ask the tomcat-devs for any ideas (although I would have a simple solution with file separator ;-) thanks klaus _________________________________________________________________ Mit MSN Fotos können Sie kinderleicht Ihre Fotos ausdrucken und Freunden zur Verfügung stellen: http://photos.msn.de -- To unsubscribe, e-mail: <mailto:[EMAIL PROTECTED]> For additional commands, e-mail: <mailto:[EMAIL PROTECTED]>