I am using netbeans to create a web service and using Glassfish as the server to test it within netbeans.
I have a file that i wish the web service to be able to read data from and possibly write to it. But where do i put the file. If 'course' is my netbeans project root i have tried placing the file in the following locations:
\Course
\Course\xml-resources\jaxb\FlightRequest
\Course\web
\Course\web\WEB-INF
\Course\src\java\org\me\FBooking
\Course\build\web
\Course\build\web\WEB-INF
\Course\build\web\WEB-INF\classes
and tried accessing it in the web service in my unmarshalling code using (as the file i am trying to access is an xml document):
un = (AvailableFlights) unmarshaller.unmarshal(new java.io.File("AvailableFlights.xml"));
But it cant find the file
So where am i supposed to place it?
If you need to write to the file, you should locate the file outside of your deployable code.
If you only need read-only access, putting the file in web/ will make the file accessible from a web browser. That may not be what you want.
If you put the file under WEB-INF/classes, it will be accessible to your code, but not publicly exposed.
Your code fragment for accessing the file will only work for files on the file system, and not for files you deploy as part of the WAR, so you need to look into other ways of loading the file if you decide to package it as part of your WAR.
Take a look at
getClass().getResourceAsStream("file")
which should be able to read from files within a WAR file. (Haven't tested it right now..) But this is only for reading from the file.
Related
I'm working with a local WebSphere server configured in IntelliJ Idea, and the application I'm working on is using a third-party library that loads a properties file with:
ThirdPartyClass.class.getClassLoader().getResourceAsStream(fileNameParameter);
It uses the default bootstrapClassLoader.
I've been instructed to make sure the properties file is in a config directory so that it can be edited without deploying a code change. My project looks something like this:
ProjectName
Configs
my.properties
src
java (sources root)
packages, .java files, etc
main (resources root)
schemas, web docs, etc
I have tried several of paths to make it work but it always returns null. Since I initially thought it was reaching from within the third party library package, I tried adding several ..\'s to the file path, but then I learned that this method loads from the classpath, so I pulled a
String test = System.getProperty("java.class.path");
and upon inspection, my classpath is all made up of websphere directories and jars within them:
C:\Users\me\Programs\IBM\AppServer\profiles\AppSrv01/properties
C:\Users\me\Programs\IBM\AppServer\AppSrv01/properties
and several jar files in C:\Users\me\Programs\IBM\AppServer/lib/
So just as a test I stuck the file in C:\Users\me\Programs\IBM\AppServer\AppSrv01/properties, then tried to grab it with just its file name (my.properties), but still couldn't reach it. I've also tried moving the file into the src directory and the main directory, but no matter what I do it just can't seem to find the file.
I'm aware that this method is typically used to grab resources from within a jar file, but from my understanding it seems like it should be possible to reach my file from outside of one as long as it's in a directory in the classpath... but apparently not since that didn't work.
I have the absolute path on my hard drive and will have said path on the server; is there a way to derive the path that ClassLoader.getResourceFromStream() wants with that info? Failing that, is there some obvious mistake I'm making with the resource url?
I think your fileNameParameter simply needs to start with / to indicate that it is in the root level of the classpath. Otherwise it will be searched relative to the class it is loaded from, i.e. the package of ThirdPartyClass in your example.
I am creating a Java Application where in I need to read templates which I am not able to. That is because I am not able to locate the files with proper path.
The code structure is as follows:
The template reading class file is in src/main/java/com.prototype.main while,
The templates are residing in src/main/java/templates
Which means, when I create a jar the three folders would be 3 folders:
com
templates
META-INF
Now from any class under the package how can I access the templates?
I have tried the following:
new File("/") which in case of web application would have picked from the root directory of the application but in stand alone applications it is the user directory of the system
System.getProperty("user.dir") which gives the project directory
new App().getClass().getName() which shows the package name separated by dot
So if I want to read a template file like /templates/some.xml, how do I get the entire path of this file?
Thanks for any help in advance.
An embedded resource is not a File, it is a resource.
You can read these resources using Class#getResource or Class#getResourceAsInputStream depending on your needs.
For example...
URL url = getClass().getResource("/templates/some.xml");
Will return a URL reference to the named resource.
You can use
InputStream i=this.getClass().getClassLoader().getResourceAsStream("templates/some.xml");
like described in xml FileNotFoundException using slick2D library in java
I am using tomcat server for my java application(Servlet,jsp).In my servlet page calling one java class function.Actually the java file is written separately and i will call the function written inside it.With in the function, i have to read one config(user defined) file for some purpose.so, i am using File class for that.
But, here i have to give relative path of the config file(user defined).Because, now i am running this application in local windows server.But my live server is based on Linux.So, the file path is changed in linux.
File f1=new File("D:\tomcat\webapp\myapp\WEB-INF\src\point_config.txt"); -- windows
File f1=new File("D:\ravi\tomcat\webapp\myapp\WEB-INF\src\point_config.txt"); -- linux
So, i have to give relative path of the file that is common to both windows and linux machine.
Is there a way to do this?
Please guide me to get out of this issue?
Place your config file under your webapp WEB-INF/classes folder and read like this in code
InputStream is=
YourClassName.class.getResourceAsStream("point_config.txt");
The path of the config file leads into the WEB-INF folder
tomcat\webapp\myapp\WEB-INF\src\point_config.txt
Anything inside WEB-INF is protected and cannot be user-defined once the web application has launched. If you meant to read from a user-defined configuration file from the file system, please use an API like the common configuration API.
If you want to insist on keeping the file inside the WEB-INF folder, use the Class.getResourceAsStream() method to obtain the configuration instead. That would not make the configuration user-defined though.
I am using one third party jar in my code. In the jar file , in one of the classes, when I opened the class using de-compiler, the code below is written:
java.net.URL fileURL = ClassLoader.getSystemResource("SOAPConfig.xml");
Now I am using this in my webapplication, where should I place this SOAPConfig.xml so that it will find the fileURL.
Note: I have tried putting this XML in WEB-INF/classes folder. But it is not working. Your help will be appreciated.
In Addition: In the explaination you have given, It is telling me not to use this code snippet inside the third party jar in this way...What is the exact usage of this statement
ClassLoader.getSystemResource will load the resource from the system classloader, which uses the classpath of the application as started from the command line. Any classloaders created by the application at runtime (i.e. the one that looks in WEB-INF/classes) are not on the system classpath.
You need to
Look through the script that starts your server, find out which directories are on the classpath there, and put your SOAPConfig.xml in one of those. If necessary, change the classpath in the script to look in a separate directory that's just used for your config file.
Track down the person who used ClassLoader.getSystemResource in the library, kick them squarely in the nuts, and tell them never to do that again.
I need to access the resource files in my web project from a class. The problem is that the paths of my development environment are different from the ones when the project is deployed.
For example, if I want to access some css files while developing I can do like this:
File file = new File("src/main/webapp/resources/styles/some.css/");
But this may not work once it's deployed because there's no src or main directories in the target folder. How could I access the files consistently?
You seem to be storing your CSS file in the classpath for some unobvious reason. The folder name src is typical as default name of Eclipse's project source folder. And that it apparently magically works as being a relative path in the File constructor (bad, bad), only confirms that you're running this in the IDE context.
This is indeed not portable.
You should not be using File's constructor. If the resource is in the classpath, you need to get it as resource from the classpath.
InputStream input = getClass().getResourceAsStream("/main/webapp/resources/styles/some.css");
// ...
Assuming that the current class is running in the same context, this will work regardless of the runtime environment.
See also:
getResourceAsStream() vs FileInputStream
Update: ah, the functional requirement is now more clear.
Actually I want to get lastModified from the file. Is it possible with InputStream? –
Use getResource() instead to obtain it as an URL. Then you can open the connection on it and request for the lastModified.
URL url = getClass().getResource("/main/webapp/resources/styles/some.css");
long lastModified = url.openConnection().getLastModified();
// ...
If what you're looking to do is open a file that's within the browser-visible part of the application, I'd suggest using ServletContext.getRealPath(...)
Thus:
File f = new File(this.getServletContext().getRealPath("relative/path/to/your/file"));
Note: if you're not within a servlet, you may have to jump through some additional hoops to get the ServletContext, but it should always be available to you in a web environment. This solution also allows you to put the .css file where the user's browser can see it, whereas putting it under /WEB-INF/ would hide the file from the user.
Put your external resources in a sub-directory of your project's WEB-INF folder. E.g., put your css resources in WEB-INF/styles and you should be able to access them as:
new File("styles/some.css");
Unless you're not using a standard WAR for deployment, in which case, you should explain your setup.
Typically resource files are placed in your war along with your class files. Thus they will be on the classpath and can be looked up via
getClass.getResource("/resources/styles/some.css")
or by opening a File as #ig0774 mentioned.
If the resource is in a directory that is not deployed in the WAR (say you need to change it without redeploying), then you can use a VM arg to define the path to your resource.
-Dresource.dir=/src/main/webapp/resources
and do a lookup via that variable to load it.
In Java web project, the standard directory like:
{WEB-ROOT} /
/WEB-INF/
/WEB-INF/lib/
/WEB-INF/classes
So, if you can get the class files path in file system dynamic,
you can get the resources file path.
you can get the path ( /WEB-INF/classes/ ) by:
this.getClass().getProtectionDomain().getCodeSource().getLocation().getPath()
so, then the next ...