[英]how to get web context path in spring mvc + maven project?
I was using tomcat7-maven-plugin to start my spring mvc project, and try to get the fullpath in spring controller by below. 我正在使用tomcat7-maven-plugin启动我的spring mvc项目,并尝试通过下面的方法获取spring控制器中的fullpath。
String path = this.getClass().getClassLoader().getResource("").getPath();
It will give me the path like below. 它将为我提供如下路径。
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/")
. 然后,我可以通过fullPath.split("/WEB-INF/classes/")
获得上下文路径。 Actually it has nothing to do with springmvc, any java web app can get context path like this if cannot get the servletContext. 实际上,它与springmvc无关,如果无法获取ServletContext,则任何Java Web应用程序都可以获取这样的上下文路径。
But if I start the project in dev mode by ' mvn tomcat7:run '. 但是如果我通过' mvn tomcat7:run '在dev模式下启动项目。 It will give me the path like below. 它将为我提供如下路径。
C:/git/xxxxx/showcase/target/classes
Then I can not get the context path by this url. 然后,我无法通过该URL获取上下文路径。 I want to know where is the context root when I start the project by maven and how can I get it? 我想知道通过Maven启动项目时上下文根在哪里,如何获得它? Thanks. 谢谢。
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;
}
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