Compile error: package javax.servlet does not exist

问题

I have a package 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.

回答1:

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.

回答2:

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>

回答3:

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

回答4:

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.

回答5:

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>

回答6:

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).

回答7:

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

回答8:

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;

回答9:

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:

回答10:

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

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.

回答11:

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.

来源：https://stackoverflow.com/questions/9193228/compile-error-package-javax-servlet-does-not-exist

标签

java

servlets

compilation

compiler-errors