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Editorial Team
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Asked: 2026-06-06T18:29:57+00:00

For testing purposes I want to use Jetty 8 to serve only static content.

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For testing purposes I want to use Jetty 8 to serve only static content. I know how to start the webserver from the command line:

java -jar start.jar jetty.port=8082

I would like to be able to use a vanilla Jetty, preferably 8 or 7, and start it using something like:

java -jar start.jar OPTIONS=resources resources.root=../foo jetty.port=8082

The files should then be accessible from the root of the server. A file called ../foo/x.html should be accessible via http://localhost:8082/x.html.

I don’t want to create a WAR file or anything fancy. Preferably it shouldn’t do any caching on the server side, leaving the files unlocked on Windows machines. Also, I only want to serve files, even located in subdirectories, no fancy file browser or ways to modify them from a client.

Is this possible? If not, what is the minimum configuration needed to accomplish such behavior?

Additional information

I’ve tried the following command. I expected to be able to browse the javadoc shipped with Jetty 8 using http://localhost:8080/javadoc/, but it always gives me a 404

java -jar start.jar –ini OPTIONS=Server,resources etc/jetty.xml contexts/javadoc.xml

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1 Answer

  1. Editorial Team

    The simplest way to start Jetty and have it serve static content is by using the following xml file:

    static-content.xml:

    <?xml version="1.0"?>
<!DOCTYPE Configure PUBLIC "-//Jetty//Configure//EN" "http://www.eclipse.org/jetty/configure.dtd">

<Configure id="FileServer" class="org.eclipse.jetty.server.Server">
    <Call name="addConnector">
      <Arg>
          <New class="org.eclipse.jetty.server.nio.SelectChannelConnector">
            <Set name="host"><Property name="jetty.host" /></Set>
            <Set name="port"><Property name="jetty.port" default="8080"/></Set>
          </New>
      </Arg>
    </Call>

    <Set name="handler">
      <New class="org.eclipse.jetty.server.handler.ResourceHandler">
        <Set name="resourceBase"><Property name="files.base" default="./"/></Set>
      </New>
    </Set>
</Configure>

    Than you can start Jetty using:

    java -jar start.jar --ini static-content.xml files.base=./foo jetty.port=8082

    If you omit files.base, the current direcory will be used; if you omit jetty.port, port 8080 will be used.

    The --ini will disable the settings from start.ini, therefore also make sure no other handlers etc. will be activated.

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