java、struts、tomcat程序中的各种取路径的方法

作者: 2011/12/9 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”;

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