Tukukkk

I was using tomcat7-maven-plugin to start my spring mvc project, and try to get the fullpath in spring controller by below.

 String path = this.getClass().getClassLoader().getResource("").getPath();

It will give me the path like below.

C:/Users/xxxxx/Documents/xxx/apache-tomcat-8.5.31/webapps/showcase/WEB-INF/classes/

Then I can get the context path by fullPath.split("/WEB-INF/classes/").
Actually it has nothing to do with springmvc, any java web app can get context path like this if cannot get the servletContext.

But if I start the project in dev mode by 'mvn tomcat7:run'. It will give me the path like below.

C:/git/xxxxx/showcase/target/classes

Then I can not get the context path by this url. I want to know where is the context root when I start the project by maven and how can I get it? Thanks.

public static File upload(MultipartFile file, HttpServletRequest request, boolean file_name, String upload_folder) {

        String filename = null;

        File serverFile = null;

        try {

            String applicationpath = request.getServletContext().getRealPath("");

            filename = file.getOriginalFilename();

            byte bytes = file.getBytes();

            String rootPath = applicationpath;

            File dir = new File(rootPath + File.separator + upload_folder);

            if (!dir.exists())

                dir.mkdirs();

            serverFile = new File(dir.getAbsolutePath() + File.separator + filename);

            BufferedOutputStream stream = new BufferedOutputStream(new FileOutputStream(serverFile));

            stream.write(bytes);

            stream.close();

            return serverFile;

        } catch (Exception e) {

            serverFile = null;

        }

        return serverFile;

    }