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I'd like to compile a very basic servlet from command prompt, but it is always unsuccessful and the compiler tells me the following:

 error: package javax.servlet does not exist.

I googled for the solution and I found that I need to include the servlet.jar libraries into my PATH. I believe I did. I strongly believe that the location for those libraries in my computer is:

C:\apache-tomcat-7.0.23\lib\servlet-api.jar\

and the end (the relevant part) of my PATH is the following:

%JAVA_HOME%\bin;C:\apache-tomcat-7.0.23\lib\servlet-api.jar\

For me, it looks ok, but it is obviously not. Can anyone tell me what could be the problem?

user
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Sanyifejű
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3 Answers3

10

classpath not path ... and you don't need it as an evironment variable. You can set the classpath for javac with option -cp or -classpath (several other ways are also available). javac uses the environment variable CLASSPATH to look for classes, that can be useful and can also be a source for hard-to-track-down-problems.

An example to compile a java file that uses a library(that is classes from outside the standard JavaSE) would be:

javac -classpath C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java

if your environmental variable CLASSPATH contains libraries you need you might want to do:

javac -classpath %CLASSPATH%;C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java

(please be aware that I don't have access to a windows machine, and therefore haven't tested the idiosyncratic parts of the syntax above) (also note that in this example "C:\apache-tomcat-7.0.23\lib\servlet-api.jar" is a jar file and not a directory which it might be from your question on your machine) For command line compiling on windows OS it is a good idea to have the environmental variable JAVA_HOME set correctly and the bin directory of the JDK in the PATH.

I do however suggest getting to write-compile-execute-deploy via/in an IDE for servlet development before figuring out how to do it with just the JDK from a command line. Java Servlets are not stand-alone executable classes but needs to be deployed into for example tomcat to be tested/used.

esej
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  • +1 for avoiding the CLASSPATH environment variable, which can lead to unexpected behavior. Better to include an example, though, for completeness. – Dave Newton Apr 28 '12 at 13:43
  • Agree an example, but I haven't seen a windows path in java in a decade or so. I'll try to update. – esej Apr 28 '12 at 14:25
  • But it'd be an example of a classpath; the only difference is the separator, `;` in Windows, `:` in real OSes. And quoting around paths with spaces, although the `\` might work too--I don't really remember Windows stuff either--which is awesome :D – Dave Newton Apr 28 '12 at 14:26
  • javac -classpath C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyServlet.java You don't need the "=" sign after -classpath. But it was a great answer, that saved my day. Thanks – Sanyifejű Apr 28 '12 at 17:06
  • Thanks (was a bluder, I blame that I usually send command line classpath alterations via -D) – esej Apr 28 '12 at 17:09
  • Hey, pr123 would you please mark this answer as the accepted answer then. This is the way stackoverflow is supposed to work :p . At the end of this faq entry is instructions how to do that. http://stackoverflow.com/faq#howtoask – esej Apr 29 '12 at 10:38
1

First copy the servlet-api.jar file from following path

C:\apache-tomcat-7.0.23\lib\servlet-api.jar;

and paste it in the bin folder of java software by following the path

C:\java\jdk1.6\bin;

Hope now you can successfully compile your servlet program.

icodebuster
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waseem
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1

1.You can copy your javax.servlet.jar in the jdk1.6\lib folder. 2.You can go to Control Panel\System\Advanced System Properties\Environment Variables

enter image description here

Enter the location of the jar in the CLASSPATH variable as follows:

enter image description here

Then compile and run the servlet.

MindBrain
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