To get the dynamic path within the project is easy, just run the code:
String path = this.getServletContext().getRealPath("");
Here it is returned:
/home/pedro-pazello/development/servers/apache-tomcat-7.0.42/wtpwebapps/img_uploader/
* img_uploader * is the name of the project. But I need to get this way out of the project. I need the path to be:
/home/pedro-pazello/development/servers/apache-tomcat-7.0.42/wtpwebapps/imagens/
Does anyone have any idea how I can get this path?
Do this:
OPTION 1
String path = this.getServletContext().getRealPath("../imagens");
File f = new File(path);
String imagensHome = f.getCanonicalPath();
You can then create a Servlet to fulfill the requests in /img_uploader and deliver the images from:
File imagensRoot = new File(imagensHome);
You should pay special attention to the Mime Type of the images at the time of delivery to the client by assigning the appropriate Content Type.
OPTION 2
Only if your server is running Tomcat 7+ on Linux, UNIX, or Mac OS X
ln -s img_uploader ../imagens
Create a context.xml file as below:
<?xml version="1.0" encoding="UTF-8"?>
<Context allowLinking="true">
<WatchedResource>WEB-INF/web.xml</WatchedResource>
</Context>
This file informs Tomcat via the allowLinking="true" attribute that it can access symbolic links and deliver content to the client from ../imagens .
NOTE: In this solution a Deploy script should re-create this symbolic link described above in case it is removed.