Get localhost:8080 instead of my_ip_address:8080 in wsdl2java

Question

I am using axistools-maven-plugin

This is my xml content.

<plugin>
    <groupId>org.codehaus.mojo</groupId>
    <artifactId>axistools-maven-plugin</artifactId>
    <version>1.4</version>
    <configuration>
    <urls>
       <url>http://localhost:8080/link/to/webservice?wsdl</url>
    </urls>
    <packageSpace>my.package</packageSpace>
    <serverSide>true</serverSide>
    <outputDirectory>src/main/java</outputDirectory>
    </configuration>
    <executions>
       <execution>
        <goals>
            <goal>wsdl2java</goal>
            </goals>
       </execution>
    </executions>
</plugin>

I get the generated output as:

http://my_ip_address:8080/link/to/webservice

I want the generated output as:

http://localhost:8080/link/to/webservice
     OR
http://127.0.0.1:8080/link/to/webservice

Because http://my_ip_address:8080/link/to/webservice (which I think) will try to get connection from network. I want to connect locally.

Please help..

UPDATE:

It works in Windows XP but not in Linux (CentOS). Sorry for not mentioning that.

score 0 · Accepted Answer · answered Dec 12 '12 at 10:05

Well, since I got no answers, I think I found my own answer.

I had to do a work around on it. Here is the process.

I download the wsdl file using the target machine itself.

Change the port address.

<service name="serviceName">
  <port name="servicePort" binding="tns:soapBinding">
    <soap:address location="http://localhost:8080/link/to/webservice"/>
  </port>
</service>

Use that wsdl file in my development machine.

Get localhost:8080 instead of my_ip_address:8080 in wsdl2java

1 Answers1