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Compile a Servlet

 
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I tried to compile a servlet like this:

javac myFirstServlet.java

It does not work, it get the following message:

package javax.servlet does not exist

How can I compile a servlet?
 
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Urs,
javax.servlet is part of Java EE not Java SE. Your server (Tomcat, JBoss, etc) comes with a file called j2ee.jar or servlet.jar. Add this to your classpath when compiling.
 
Urs Waefler
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I know, that javax.servlet is not part of Java SE. I use Apache Tomcat. I found servlet.jar. So far so good.

How shall I continue?

This is the path:

C:\Programme\Apache Software Foundation\Tomcat 6.0\lib\servlet.jar

javac -classpath C:\Programme\Apache Software Foundation\Tomcat 6.0\lib\servlet.jar myFirstServlet.java

Is it something like that? Or do I have to copy servlet.jar in folder lib of the jdk?

Till now it does not work.
 
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Do not copy the jar file anywhere. You can include it in the classpath in-place. Otherwise you risk jars getting out of sync with each other.
 
Urs Waefler
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Now it worked, I could compile it. I did it like that:

set classpath=C:\Programme\Apache Software Foundation\Tomcat 6.0\lib\servlet-api.jar

Then:

javac myFirstServlet.java

But I do not want to enter the first line always. Is it possible to set it in the operating system, in Microsoft Windows XP you can set in System the variables.
 
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Use the CLASSPATH variable.
 
Greenhorn
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You can set the CLASSPATH as either a user variable or system variable. As the name says, if it is set as a user variable, it is available for the particular user. If it is set as a system variable, the classpath is available to all.

 
Java Cowboy
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See this entry in the FAQ: Compiling Servlets
 
Urs Waefler
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Thank you, now it works fine. I set the environment variable CLASSPATH.
 
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