I have several swagger yaml files and I want to generate code at compile time using maven.
here is my code:
<plugins>
<plugin>
<groupId>io.swagger</groupId>
<artifactId>swagger-codegen-maven-plugin</artifactId>
<version>2.3.1</version>
<executions>
<execution>
<goals>
<goal>generate</goal>
</goals>
<configuration>
<inputSpec>${project.basedir}/src/main/resources/*.yaml</inputSpec>
<language>java</language>
<configOptions>
<sourceFolder>src/gen/java/main</sourceFolder>
</configOptions>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
I'm trying to get all the yaml files I have in the resources as input :
<inputSpec>${project.basedir}/src/main/resources/*.yaml</inputSpec>
but this do not work, I've got this error:
failed to read resource listing
java.io.FileNotFoundException: C:\Projets\www\codegen\src\main\resources\*.yaml (La syntaxe du nom de fichier, de répertoire ou de volume est incorrecte)
Any Idea ?
My problem was resolved by using multiple executions as follow :
<plugin>
<groupId>io.swagger</groupId>
<artifactId>swagger-codegen-maven-plugin</artifactId>
<version>2.4.5</version>
<executions>
<execution>
<id>a</id> .....
</execution>
<execution>
<id>b</id> .....
</execution>
<execution>
<id>c</id> .....
</execution>
</executions>
</plugin>