I have a little problem with Struts 2 when I try to get the context path :
ServletActionContext.getServletContext().getRealPath("\\WebContent\\resources\\img\\");
I got this path:
C:\Users\killian\workspace.metadata.plugins\org.eclipse.wst.server.core\tmp0\wtpwebapps\SiteWebAdministrable\WebContent\resources\imgicone.jpg
Why the exact source path ?
Because i need to upload and save images for an admin website to control background and without the actual path i cannot save images in the resources path...
So i save the path with the name and extension in the database (no problem), and i need to save the image in the resource directory (image problem...)
Can someone help me please ? Did i forgot something ?
This question is the answer ?
How do you get the project path in Struts 2?
servletContext.getServletContext().getRealPath("/resources/img/name_of_image.png")
So, passing the "/" to getRealPath() would return you the absolute disk file system path of the /web folder of the expanded WAR file of the project. Something like /path/to/server/work/folder/demo.war/ which you should be able to further use in File or FileInputStream.
Note that most starters don't seem to see/realize that you can actually pass the whole web content path to it and that they often use
String absolutePathToIndexJSP = servletContext.getRealPath("/") + "demo.png";
instead of
String absolutePathToIndexJSP = servletContext.getRealPath("/demo.png");
getRealPath() is unportable; you'd better never use it
Use getRealPath() carefully.
If all you actually need is to get an InputStream of the web resource, better use ServletContext#getResourceAsStream() instead, this will work regardless of the way how the WAR is expanded. So, if you for example want an InputStream of index.jsp, then do not do:
InputStream input = new FileInputStream(servletContext.getRealPath("/demo.png")); // Wrong!
But instead do:
InputStream input = servletContext.getResourceAsStream("/demo.png"); // Right!
Or if you intend to obtain a list of all available web resource paths, use ServletContext#getResourcePaths() instead.
Set<String> resourcePaths = servletContext.getResourcePaths("/");
Related
I know there are many topics out there for this but I have seem to have tried everything. I can put my file name in there and it finds it if there is a src folder,
InputStream is = context.class.getClassLoader().getResourceAsStream("file.props");
but when we put it on an apache server, a src folder is not automatically created, so it isn't finding it. I have tried placing it directly in the web-inf folder and
InputStream is = context.class.getClassLoader().getResourceAsStream("/WEB-INF" + File.separator + "file.props");
But this is always returned as null. What is the reason for this? The file exists there, why can't it find it?
You appear to be using the wrong ClassLoader. Invoking context.class.getClassLoader() provides the ClassLoader with which the ServletContext class (context.class) was loaded. What you want is the ClassLoader for the web application's classes, which would be context.getClassLoader().
Don't use the ClassLoader if you want to load your file from /WEB-INF. Instead, use the ServletContext's method for just that purpose:
// In your servlet e.g. doGet method
ServletContext app = super.getServletContext();
InputStream in = app.getResourceAsStream("/WEB-INF/file.props");
Note that using / is okay regardless of the OS, filesystem, etc.
If you really want to use the ClassLoader, take #rickz's advice and move your file.props into WEB-INF/classes.
The code I am using to load the image is:
ImageIO.read(SpriteSheet.class.getResource(path));
The path being the path to the resource. But it would error with IllegalArgumentException. I wondered what might be causing and came to the conclusion that the resource should be added into the same path as the class.
Is it possible to load the image from another folder, like a res folder outside of the bin folder? (folder holding compiled classes)
EDIT:
So i messed around with a few things, and came to a solution. But now I have another problem. Here is my code
File sheet = new File(SpriteSheet.class.getProtectionDomain().getCodeSource().getLocation().getPath());
URI uri = sheet.toURI();
BufferedImage image = ImageIO.read(uri.toURL());
When I try to run it, it gives me an IIOException: Can't read Input File
This means that I can never actually get it work. I tried debugging by prining the URL to the console and this is the URL.
C:\Users\Amma\Abhijeet\Eclipse%20Workspace1\Test%20Game\bin
The %20 comes in the middle. Meaning that the file is and never can be acceesed. Is there anyway I can fix this?
Thanks.
Class.getResource will return null if the resource could not be found or the invoker doesn't have adequate privileges to get the resource.
All variants of ImageIO.read will throw an IllegalArgumentException if they receive a null input.
Take a look at the documentation of the getResource to understand how an absolute resource name is constructed from the given resource named and what are the rules for searching resources.
You can read images from any location as long as you have permissions to do so, the ImageIO.read method accepts a File, URL or InputStream so you have many option to do it.
I have a java app where I'm trying to load a text file that will be included in the jar.
When I do getClass().getResource("/a/b/c/"), it's able to create the URL for that path and I can print it out and everything looks fine.
However, if I try getClass().getResource(/a/b/../"), then I get a null URL back.
It seems to not like the .. in the path. Anyone see what I'm doing wrong? I can post more code if it would be helpful.
The normalize() methods (there are four of them) in the FilenameUtils class could help you. It's in the Apache Commons IO library.
final String name = "/a/b/../";
final String normalizedName = FilenameUtils.normalize(name, true); // "/a/"
getClass().getResource(normalizedName);
The path you specify in getResource() is not a file system path and can not be resolved canonically in the same way as paths are resolved by File object (and its ilk). Can I take it that you are trying to read a resource relative to another path?
This question already has answers here:
How to find the working folder of a servlet based application in order to load resources
(3 answers)
Closed 6 years ago.
I currently have a bunch of images in my .war file like this.
WAR-ROOT
-WEB-INF
-IMAGES
-image1.jpg
-image2.jpg
-index.html
When I generate html via my servlets/jsp/etc I can simple link to
http://host/contextroot/IMAGES/image1.jpg
and
http://host/contextroot/IMAGES/image1.jpg
Not I am writing a servlet that needs to get a filesystem reference to these images (to render out a composite .pdf file in this case). Does anybody have a suggestion for how to get a filesystem reference to files placed in the war similar to how this is?
Is it perhaps a url I grab on servlet initialization? I could obviously have a properties file that explicitly points to the installed directory but I would like to avoid additional configs.
If you can guarantee that the WAR is expanded, then you can use ServletContext#getRealPath() to convert a relative web path to an absolute disk file system which you can further use in the usual Java IO stuff.
String relativeWebPath = "/IMAGES/image1.jpg";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath);
InputStream input = new FileInputStream(file);
// ...
However, if you can't guarantee that the WAR is expanded (i.e. all resources are still packaged inside WAR) and you're actually not interested on the absolute disk file system path and all you actually need is just an InputStream out of it, then use getServletContext().getResourceAsStream() instead.
String relativeWebPath = "/IMAGES/image1.jpg";
InputStream input = getServletContext().getResourceAsStream(relativeWebPath);
// ...
See also:
getResourceAsStream() vs FileInputStream
Use the getRealPath method of ServletContext.
Ex:
String path = getServletContext().getRealPath("WEB-INF/static/img/myfile.jpeg");
This is relatively straight forward you simply use the class loader to fetch the files from the class plath. :
InputStream is = YourServlet.class.getClassLoader().getResourceAsStream("IMAGES/img1.jpg");
There are a few other getResoruce classes that are worth looking at. Also you don't have to fetch the class loader through the class variable on your servlet. Any class that you happen to know has been loaded by the container should work .
If you know the relative location of the files you could ask the runtime about the exact location using
Thread.currentThread().getContextClassLoader().getResource(<relative-path>/<filename>)
This would give you an URL to the location where the specified image can be found. This URL can be used to read the specified file or you can split it to use the different parts of the URL for further processing.
I'm trying to upload an external image and I need to save it in a folder where I have a managed bean. Any idea how I could do this?
Use Class#getResource() to obtain the URL where the class is located.
URL beanClassPath = Bean.class.getResource("");
File imageFile = new File(beanClassPath.getPath(), imageFileName);
// ...
However, this is generally a very bad idea. If you redeploy the webapp, everything will get lost. Rather store the images in a fixed path somewhere outside the webapplication, e.g. /images or so. In a Windows environment this will automatically refer to the disk from where the webapplication is started, e.g. c:/images.
File imageFile = new File("/images", imageFileName);
// ...
You can also consider to store them in a database, you'll only need to store some metadata along it, such as the original filename, content type and preferably also the content length and eventually the creation and last modification timestamps. That kind of information which you would usually obtain using the java.io.File methods. You'll namely going to need them whenever you'd like to serve the image back to the webpage.