title: How do I change the API name in the class generated by swagger-codegen
body: <p>I am using <a href="https://github.com/martypitt/swagger-springmvc" rel="noreferrer">swagger-springmvc</a> and <a href="https://github.com/swagger-api/swagger-codegen" rel="noreferrer">swagger-codegen</a> to generate a Java client library for a RESTful webservice. I have written my own extension of the BasicJavaGenerator (see below) to override the package names, and can successfully generate the client library files. The "main" files generated are:</p>
  
  <pre><code>swagger-codegen/generated-code/java/pom.xml
  swagger-codegen/generated-code/java/src/main/java/com/example/ApiApi.java
  swagger-codegen/generated-code/java/src/main/java/com/example/model/*.java
  </code></pre>
  
  <p>What I can't find is how to set the name of my API to get the code generator to rename ApiApi.java to MyProjectApi.java (for example) as seems to be done in the samples included in with swagger-codegen. I have tried looking at the code generator code to override the api name, and have also tried looking both at the swagger json spec and swagger springmvc functionality to an option to set the name.</p>
  
  <p>The code generator:</p>
  
  <pre class="lang-scala prettyprint-override"><code>package com.wordnik.swagger.codegen
  
  import com.wordnik.swagger.codegen.BasicJavaGenerator
  
  object MyJavaGenerator extends BasicJavaGenerator {
    def main(args: Array[String]) = generateClient(args)
  
    // api invoker package
    override def invokerPackage = Some("com.example.api")
  
    // package for models
    override def modelPackage = Some("com.example.api.model")
  
    // package for api classes
    override def apiPackage = Some("com.example.api")
  }
  </code></pre>