java、struts、tomcat程序中的各种取路径的方法

东坡下载 2011年12月09日 10:56:22

      我和一向对程序比较感兴趣,但因为很多原因没有能坚持下来,不过也收集了一些有用的代码吧.

      得到当前路径
      File f = new File(“.”);
      String absolutePath = f.getAbsolutePath();
      System.out.println(absolutePath);

      1.路径中空格:

      String pathtem=null;
      URL ut=Thread.currentThread().getContextClassLoader().getResource(“”);
      pathtem=ut.toString();
      pathtem=pathtem.replace(“%20″, ” “);//去掉空格
      pathtem=pathtem.substring(6,pathtem.length());//去掉”file:/”
      返回:
      d:/apache/webapp/Ev/WEB-INF/classes/

      struts中得到路径
      String rootuploadpath = this.getServlet().getServletContext().getRealPath(“/”);

      2.生成jar包后
      String pathtem=null;
      URL ut=Csssb_se.class.getResource(“”);
      pathtem=ut.toString();
      pathtem=pathtem.replace(“%20″, ” “);
      pathtem=pathtem.substring(0,pathtem.indexOf(projectname));
      pathtem=pathtem.replace(“file:///”,”");
      pathtem=pathtem.replace(“file:/”,”");
      pathtem=pathtem.replace(“jar:”,”");

      3.生成war包后,Tomcat

      String Dirpath = null;
      String Project_name = “TranData”;
      URL ut = Thread.currentThread().getContextClassLoader().getResource(“”);
      Dirpath = ut.toString();
      Dirpath = Dirpath.replace(“%20″, ” “);
      Dirpath = Dirpath.substring(6, Dirpath.length());
      int project_name_index = Dirpath.indexOf(Project_name);
      Dirpath = Dirpath.substring(0, project_name_index
      + Project_name.length() + 1);

      String xmlfilepath = Dirpath + “s.xml”;