javac command line compile error: package javax.servlet does not exist

Question

I have a Servlet class in which I import javax.servlet.* and javax.servlet.http.*. When I try to compile it in command prompt I get the error

package javax.servlet does not exist

I use JDK 1.7.0 and Tomcat 6.0. I compile using javac. I am not using a build tool like Maven.

BalusC · Accepted Answer · 2021-12-02T11:59:11.270

102

You need to add the path to Tomcat's /lib/servlet-api.jar file to the compile time classpath.

javac -cp .;/path/to/Tomcat/lib/servlet-api.jar com/example/MyServletClass.java

The classpath is where Java needs to look for imported dependencies. It will otherwise default to the current folder which is included as . in the above example. The ; is the path separator for Windows; if you're using an Unix based OS, then you need to use : instead.

If you're still facing the same complation error, and you're actually using Tomcat 10 or newer, then you should be migrating the imports in your source code from javax.* to jakarta.*.

import jakarta.servlet.*;
import jakarta.servlet.http.*;

In case you want to keep using javax.* for whatever reason, then you should be downgrading to Tomcat 9 or older as that was the latest version still using the old javax.* namespace.

2

true, but in the future consider using maven to solve this kind of problems for you – Kris Feb 08 '12 at 12:44
12

Or just an IDE like Eclipse. I'd however recommend continuing learning the hard way until you can almost dream it. Otherwise it will be hard to understand how IDEs work under the covers. – BalusC Feb 08 '12 at 12:46
My apology, if I have more than one servlet class, how can I add those to resolve this problem? thanks – Stu_Dent Aug 30 '20 at 05:30
@Stu_Dent: [how to compile multiple files using javac](https://www.google.com/search?q=how+to+compile+multiple+files+using+javac) – BalusC Aug 30 '20 at 09:07

score 42 · Answer 2 · answered Nov 09 '15 at 05:11

42

If you are working with maven project, then add following dependency to your pom.xml

<dependency>
    <groupId>javax.servlet</groupId>
    <artifactId>javax.servlet-api</artifactId>
    <version>3.0.1</version>
    <scope>provided</scope>
</dependency>

answered Nov 09 '15 at 05:11

Swalih

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10
10

2

The Gradle equivalent is `dependencies { compile group: 'javax.servlet', name: 'servlet-api', version:'2.4' }` – MarkHu Dec 15 '16 at 01:47
Worked like a charm. Thanks. – raksheetbhat May 23 '18 at 12:39

score 9 · Answer 3 · edited Sep 25 '15 at 08:12

Is it a JSP or Servlet?

Well, these two packages aren’t actually built into Java like java.io is. Instead, they come with the Servlet-capable Web server (e.g. Tomcat). So before the Java compiler will be able to compile our Servlet, we need to let it know where to find the classes in these two packages.

The classes required are normally stored in a file called servlet.jar. The exact location of this file will depend on the particular Web server software you use, but in the case of Tomcat you can find it in the lib subdirectory of the main Tomcat installation directory (e.g. d:\Program Files\Apache Group\jakarta-tomcat-3.2.3\lib\servlet.jar). For the Java compiler to be able to compile Servlets, you need to add this file to your Java class path. By default, Java looks for classes in the current directory (".") only. Thus, "." is the default class path. If you change the class path to include the servlet.jar file (".;d:...\lib\servlet.jar" under Windows, ".:/usr/.../lib/servlet.jar" in Unix), then the Servlet should compile just fine.

You can specify a class path to use when you run javac.exe as follows:

d:\javadev> javac -classpath ".;d:\Program Files\Apache Group\ jakarta-tomcat-3.2.3\lib\servlet.jar" MyServlet.java

Or in Linux javac uses : instead of ;

server1> javac -classpath ".:./servlet/servlet.jar" MyServlet.java

should I do this anytime I compile a file? I mean is there a way to set classpath for all the time I compile a file. — Karadous, Feb 08 '12 at 12:46
For you to compile the file i.e. *.java, you have to ensure the servlet.jar is in the classpath. Note that JSPs eventually get translated to servlets which are, ofcourse, Java files. — Frankline, Feb 08 '12 at 13:22
Someone has the advice: *"If you are using Windows Adding d:\Program Files\Apache Group\ jakarta-tomcat-3.2.3\lib\servlet.jar; to JAVA_HOME Variable also does the magic"* — Pubby, Apr 29 '12 at 16:21

score 5 · Answer 4 · answered May 27 '13 at 17:53

In a linux environment the soft link apparently does not work. you must use the physical path. for instance on my machine I have a softlink at /usr/share/tomacat7/lib/servlet-api.jar and using this as my classpath argument led to a failed compile with the same error. instead I had to use /usr/share/java/tomcat-servlet-api-3.0.jar which is the file that the soft link pointed to.

score 4 · Answer 5 · answered Aug 05 '16 at 17:02

4

This is what solved the problem for me:

<dependency>
    <groupId>javax.servlet.jsp</groupId>
    <artifactId>jsp-api</artifactId>
    <version>2.2</version>
    <scope>provided</scope>
</dependency>

answered Aug 05 '16 at 17:02

Koray Tugay

22,894
45
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score 4 · Answer 6 · answered Apr 15 '18 at 12:15

4

Copy the file "servlet-api.jar" from location YOUR_INSTILLATION_PATH\tomcat\lib\servlet-api.jar and Paste the file into your Java Directory YOUR_INSTILLATION_PATH\Java\jdk1.8.0_121\jre\lib\ext

this will work (tested).

answered Apr 15 '18 at 12:15

Md Rehan

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2
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score 1 · Answer 7 · edited Apr 19 '17 at 18:39

JSP and Servlet are Server side Programming. As it comes as an built in package inside a Server like Tomcat. The path may be like wise

C:\Program Files\Apache Software Foundation\Tomcat 6.0\lib\jsp-api.jar
C:\Program Files\Apache Software Foundation\Tomcat 6.0\lib\servlet-api.jar

Just you want to Do is Add this in the following way

Right Click> My Computer>Advanced>Environment Variables>System variables

Do> New..> Variable name:CLASSPATH
           Variable value:CLASSPATH=.;C:\Program Files\Apache Software Foundation\Tomcat 6.0\lib\servlet-api.jar;

score 1 · Answer 8 · answered Jun 29 '20 at 17:33

This happens because java does not provide with Servlet-api.jar to import directly, so you need to import it externally like from Tomcat , for this we need to provide the classpath of lib folder from which we will be importing the Servlet and it's related Classes.

For Windows you can apply this method:

open command prompt
type

 javac -classpath "C:\Program Files\Apache Software Foundation\Tomcat 9.0\lib\*;" YourFileName.java

It will take all jar files which needed for importing Servlet, HttpServlet ,etc and compile your java file.
You can add multiple classpaths Eg.

javac -classpath "C:\Users\Project1\WEB-INF\lib\*; C:\Program Files\Apache Software Foundation\Tomcat 9.0\lib\*;" YourFileName.java

score 1 · Answer 9 · answered Feb 21 '23 at 18:46

Now that we are transitioning to from javax to jakarta namespace. I added a maven dependency

        <dependency>
        <groupId>jakarta.servlet</groupId>
        <artifactId>jakarta.servlet-api</artifactId>
        <version>6.0.0</version>
        <scope>provided</scope>
    </dependency>

and then edited the file

nbfs://nbhost/SystemFileSystem/Templates/JSP_Servlet/ServletListener.java

by changing the import

import javax.servlet.annotation.WebListener;

to

import jakarta.servlet.annotation.WebListener; now all my listener classes are created pre-jakartified.

I hope that helps cheers!

Yep, this is the right answer. Took me like the whole day to figure it out but it is because `javax` is being replaced with `jakarta`. More details [here](https://stackoverflow.com/questions/73350585/upgrade-from-spring-boot-2-7-2-to-spring-boot-3-0-0-snapshot-java-package-java/75786362#75786362) if anyone's curious. — yuyu5, Mar 20 '23 at 03:38

score 1 · Answer 10 · answered Feb 08 '12 at 12:38

1

Add servlet-api.jar into your classpath. It will be available into Tomcat's lib folder.

answered Feb 08 '12 at 12:38

Abhendra Singh

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Tiina · Answer 11 · 2021-06-02T01:03:50.680

This is what I found. Adding /usr/local/apache-tomcat-7.0.64/lib/servlet-api.jar in my environment variable as CLASSPATH on Mac.

if using bash: ~/.bash_profile $CLASSPATH=/usr/local/apache-tomcat-7.0.64/lib/servlet-api.jar

if using zsh: ~/.zshrc export CLASSPATH="usr/local/apache-tomcat-7.0.64/lib/servlet-api.jar"

Force it work right now, run source .bash_profile (or .zshrc) or one can restart computer and it works for the current user.

score 0 · Answer 12 · answered Mar 19 '19 at 02:50

Even after trying suggested solution, it was not solving my problem because there where many instance of java path were entered by me.

I removed all java related path (different version java) from "Path, JAVA_HOME, JRE_HOME" and created from fresh.
i have set (path may changes as per different installation)
a. JAVA_HOME as C:\Program Files\Java\jdk1.8.0_191
b. JRE_HOME as C:\Program Files\Java\jdk1.8.0_191\jre\lib
c. add binary file path in path: C:\Program Files\Java\jdk1.8.0_191\bin
d. CLASSPATH as C:\apache-tomcat-7.0.93\lib
never try in the same command prompt if its already open while doing changes/creting system/user variables. close it and open new one.

Reference Image:

score 0 · Answer 13 · answered Apr 09 '21 at 05:28

http://www.java2s.com/Code/JarDownload/javax.servlet/javax.servlet.jar.zip

Download zip file from the location and place it into following path after unziping files

%JAVA_HOME%/jre/lib/ext/

place files in the without any directory

score -2 · Answer 14 · edited Apr 07 '19 at 17:16

-2

The possible solution (Tested on ubuntu)

Open Terminal type geany .bashrc
Go to the top and paste this
export CLASSPATH=$CLASSPATH:/web/apache-tomcat-8.5.39/lib/servlet-api.jar
Now save and close
Try running the program now.

edited Apr 07 '19 at 17:16

Gahan

4,075
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answered Apr 07 '19 at 16:40

sadhana

1

javac command line compile error: package javax.servlet does not exist

14 Answers14

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