Setting Classpath using command prompt in Windows 7

Question

I am compiling my code through Windows 7 using command prompt -- here are details :

I set the class path like this :

set classpath= %classpath%;C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\*.jar;C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet\*.jar;

and then I tried to compile my file like :

javac -d ..\classe com\java\controller\LoginServlet.java

output:

com\java\controller\LoginServlet.java:7: package javax.servlet does not exist import javax.servlet.RequestDispatcher; ^ com\java\controller\LoginServlet.java:8: package javax.servlet does not exist import javax.servlet.ServletException; ^ com\java\controller\LoginServlet.java:9: package javax.servlet.http does not exist import javax.servlet.http.HttpServlet; ^ com\java\controller\LoginServlet.java:10: package javax.servlet.http does not exist import javax.servlet.http.HttpServletRequest; ^ com\java\controller\LoginServlet.java:11: package javax.servlet.http does not exist import javax.servlet.http.HttpServletResponse; ^ com\java\controller\LoginServlet.java:13: cannot find symbol symbol: class HttpServlet public class LoginServlet extends HttpServlet{ ^ com\java\controller\LoginServlet.java:21: cannot find symbol symbol : class HttpServletRequest location: class com.java.controller.LoginServlet public void service(HttpServletRequest request, HttpServletResponse response) ^ com\java\controller\LoginServlet.java:21: cannot find symbol symbol : class HttpServletResponse location: class com.java.controller.LoginServlet public void service(HttpServletRequest request, HttpServletResponse response) ^ com\java\controller\LoginServlet.java:22: cannot find symbol symbol : class ServletException location: class com.java.controller.LoginServlet throws ServletException, IOException { ^ com\java\controller\LoginServlet.java:46: cannot find symbol symbol : class RequestDispatcher location: class com.java.controller.LoginServlet RequestDispatcher dispatcher = request.getRequestDispatcher("/home.jsp"); ^ com\java\controller\LoginServlet.java:20: method does not override or implement a method from a supertype @Override

after that I tried like :

javac -classpath C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\*.jar com\java\controller\LoginServlet.java

then the output I got is :

javac: invalid flag: C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\servlet-api-2.5.jar Usage: javac <options> <source files> use -help for a list of possible options

Please Help on this As I am stuck on this point and I am not getting anything..how to go forward.I need help badly :(

Thanks in Advance

Possible duplicate: stackoverflow.com/questions/1237093/…. Try removing the .jar. — Ray Stojonic
– Ray Stojonic, Commented Sep 14, 2013 at 13:20

Gyanendra Dwivedi · Accepted Answer · 2013-09-14 14:48:57Z

Understanding class path wildcards

Class path entries can contain the basename wildcard character , which is considered equivalent to specifying a list of all the files in the directory with the extension .jar or .JAR. For example, the class path entry foo/ specifies all JAR files in the directory named foo. A classpath entry consisting simply of * expands to a list of all the jar files in the current directory.

A class path entry that contains * will not match class files. To match both classes and JAR files in a single directory foo, use either foo;foo/* or foo/*;foo. The order chosen determines whether the classes and resources in foo are loaded before JAR files in foo, or vice versa.

Subdirectories are not searched recursively. For example, foo/* looks for JAR files only in foo, not in foo/bar, foo/baz, etc.

Must read full detail here, it is awesome http://docs.oracle.com/javase/6/docs/technotes/tools/windows/classpath.html

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