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Editorial Team
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Asked: 2026-05-27T04:13:57+00:00

I created simple Java Servlet: WelcomeServlet.java. Than, I tried compile this file via: javac

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I created simple Java Servlet: WelcomeServlet.java.

Than, I tried compile this file via:

javac WelcomeServlet.java

In result I see compile error:

package javax.servlet doesn’t exit

I try find solution for this error with Google. And I find first part of answer: java compiler doesnt see servlet-api.jar file.

I know, that Apache Tomcat in it lib folder contains servlet-api.jar file.

So, I have this file, but where I must copy this file??

I try different folders:

echo %JAVA_HOME%
C:\Program Files\Java\jdk1.6.0_26

%PATH% contains this line: C:\Program Files\Java\jdk1.6.0_26\bin

So, I copy in:
%JAVA_HOME%\bin
%JAVA_HOME%\lib
%JAVA_HOME%\jre\lib

And in result same error.
And only after I copy servlet-api.jar in directory:
%JAVA_HOME%\jre\lib\ext

compilation complite sucessful.

My question: Why? Why I must copy in folder %JAVA_HOME%\jre\lib\ext ??

Where This moment describe in documentation?

And other question we have some official docs or specifications that describe folder structure for jdk folder??

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1 Answer

  1. Editorial Team

    You’ll need to specify the directory or directories you want the compiler to search by using the -classpath command line option when running javac. The reason the compiler found your .jar in %JAVA_HOME%\jre\lib\ext is because it searches the extension directories by default.

    This is for Java 1.5, but I believe it is more or less still correct:

    http://docs.oracle.com/javase/1.5.0/docs/tooldocs/findingclasses.html

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