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.
Related
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("/");
I've seen some posts on using java.lang.Class.getResources() and java.lang.Class.getResourcesAsStream() on SO today. Somehow, I still have confusion.
I have a Jar file that contains this structure (resources/test.xml)
This jar file is on the classpath of my application, and when I call below piece of code, it returns null, i.e. value of mappingURL is null.
URL mappingURL = this.getClass().getResource("/resources/test.xml");
However when I store the XML file in exploded format on the classpath i.e. by creating a directory "resources" and storing mapping.xml inside, it works.
I'm using this URL for reading the content of the "test.xml" file later.
Does that mean, getResources() is not the appropriate method for reading the files from inside a Jar? I didn't understand why mappingURL is null when file (test.xml) is present in the Jar file?
The getResource() method will return null if it cannot find the resource. You are prefixing the resource with a / which means that it is trying to look in the folder. You should be able to remove the leading / and achieve your intended result.
Here is the getResource() method description from the documentation:
Finds the resource with the given name. A resource is some data (images, audio, text, etc) that can be accessed by class code in a way that is independent of the location of the code.
The name of a resource is a '/'-separated path name that identifies the resource.
This method will first search the parent class loader for the resource; if the parent is null the path of the class loader built-in to the virtual machine is searched. That failing, this method will invoke findResource(String) to find the resource.
this.getClass().getResource("/resources/test.xml");
That should work, provided that the class file you are starting with is also in the same JAR file.
Does that mean, getResources() is not the appropriate method for
reading the files from inside a Jar?
This is correct. If your resource will be bundled in a jar, you always want to use getResourceAsStream().
A single line later and you can start reading the file:
BufferedReader reader = new BufferReader(InputStreamReader(getResourceAsStream("/test.xml")));
reader.readLine();
//...
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 load a properties file without using the actual path of the file. I've already done that on some other simple apps using:
InputStream inputStream = ClassLoader.getSystemResourceAsStream(PROPERTIES_FILE);
props.load(inputStream);
But this time it doesn't work. The inputStream is null for some reason. PROPERTIES_FILE is a constant defined as "app.properties". I tried to remove the .properties extension and got the same results.
Any ideas?
Thanks.
The PROPERTIES_FILE constant should include the package as well as the properties file (e.g. "com/some/library/file.properties".
final static String PROPS_FILE = "/com/some/library/file.props";
//The preceding "/" is dependendant on wheterh
//you are going to be giving a relative or absolute location
InputStream is = YourCurrentClass.class.getResourceAsStream(PROPS_FILE);
Got the same problem.
Reason: I renamed DAO package to dao. While exploding the artifact, directory DAO not get overwritten.
So I got dao in project internals and DAO in filesystem :facepalm:
When getSystemResourceAsStream returns null, it means the resource was not found. Make sure the requested resource really is on the classpath at the specified location.