I have current working file's directory path I:\\apache-tomcat-7.0.40\\webapps\\ExecutableFileProcess\\WEB-INF\\classes\\PackageName\\
I want path like I:\\apache-tomcat-7.0.40\\webapps\\ExecutableFileProcess\\
. How can I do this in java code?
You better get your App-Path from the ServletContext
in Servlet-Containern like Tomcat.
Use
servletContext.getRealPath(".");
// will return 'I:\apache-tomcat-7.0.40\webapps\ExecutableFileProcess\' on win.
// will return '/usr/local/tomcat/webapps/ExecutableFileProcess' on unix.
Have a look at
http://developer.android.com/reference/java/io/File.html#getParentFile()
Applying this twice will probably get you the right directory.
But note that you program might or might not be allowed to go there, based on the settings in the file system and the context your code runs in.
Try this:
Path aPath = Paths.get("I:\\apache-tomcat-7.0.40\\webapps\\ExecutableFileProcess\\WEB-INF\\classes\PackageName\\");
Path parentsParentpath = aPath.getParent().getParent();
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