I have deployed my web application in GlassFish server 3.0. When I execute it, I get an error: The system cannot find the file specified (MobileOntologyRev1.owl) , which is a file from which I read in my code (I haven't specified the absolute path for this file in my code, and simply refer it using the file name without any addtional path) . Where should this file be kept in order to access it? I have presently tried keeping it inside the WEB-INF/Classes folder and in the root dir of the application inside glassfish/domains/domain1/
Where should I place this file??
You may consider taking advantage of the FaceContext as mentioned below.
You can create a folder (repors) inside your WEB-INF for example.
String pathToFile=
FacesContext.getCurrentInstance().getExternalContext().getRealPath("/WEB-INF/reports/MobileOntologyRev1.owl");
Related
I have created the default play application in IntelliJ in directory P. I have over-written the default index.scala.html with my own html code. The html code refers to some css and js files which are outside the directory P. To include these external files, I added the directory of these files using project configuration settings.
My webpage doesn't load properly as the server returns 404 for the css and js files. What am I doing wrong?
When you added your directory using project structure, you only say:
Hey, IDEA, please consider this folder part of my project, consider
its contents source code and display it when I open my project.
However, when you deploy or run your app, you only deploy the usual folders to the server, which contain the resources which will be available for clients to access.
The external directory is not part of these directories and will not be deployed.
What you can do is to copy the file from the external directory as a part of your build process before deploying the application.
EDIT: Detailed answer here: What is intellij's build process for play applications
I have a net core app now running on full net452 framework. I have included a file (test.txt) in wwwroot that needs to be accessed from the net452 libraries. I have injected the hosting environment and gotten the path. However, System.IO.File.Exists(path\filename) returns false and the dll code returns "Specified file does not exist." The file does exist and the path "C:\Users\bbrown\Source\Repos\MyRepo\wwwroot\test.txt" looks correct. I am 99% certain that it is a permissions problem. A straight net452 app with path of "C:\Users\bbrown\Documents\TestApp\AppData\test.txt" works.
How can I gain access to the file or at least debug the problem.
Thanks!
1) Make sure you have proper permission for that folder. Go to that folder under security and see if you have permission or not. If you are using IIS then make sure that your application pool user has a permission on that folder.
2) _hostingEnvironment.ContentRootPath will give you the content root path which means the folder itself. You should use _hostingEnvironment.WebRootPath which will give you the path of wwwroot folder.
Here is the nice blog that will clear your doubts
https://blog.mariusschulz.com/2016/05/22/getting-the-web-root-path-and-the-content-root-path-in-asp-net-core
I have written a java Servlet web application, using NetBeans 7.2.1. The program have some jar file libraries that I have attached to the project. The application runs fine using NetBeans and Apache Tomcat 7.0.27.
My problem is that some of the jar file libraries that I am using in the project, need to access to some folders and files. I put these folder and files on the same directory as the whole NetBeans project is. but I got this exception:
Exception: java.lang.RuntimeException: java.io.FileNotFoundException
So I used these codes to find out where should I put them:
out.println("current directory: " + new File(".").getAbsolutePath());
out.println("current directory: " + System.getProperty("user.dir"));
out.println("current directory: " + getServletContext().getRealPath(("/")));
So I figured out that the current working directory is:
C:\Program Files\Apache Software Foundation\Apache Tomcat 7.0.27\bin
My question is that how can I set different directory address for each web application? I have many web applications and some of them use the same resource file names. I can't just put all of them in one directory.
Please note that I don't have access to the source code of jar files to change the. I just need a way to set the absolute path that the jar files use.
I have the same problem when I put the WAR file on the unix server. The extracted WAR file is in this location on the server:
/data02/tools/Apache/Tomcat/apache-tomcat-6.0.37/webapps/BANNEROnline
But I figure I should put the resource folders and files in this path (moosavi3 is my username!):
/home/moosavi3
How can I change the path?
The working directory is the directory from which java.[exe,bin] is started. I assume the bin directory is where the tomcat start-up script is? If the jars are all using this working directory I don't believe there is a way to make different web-apps have different working directory, they're all loaded on the same jvm (java.exe) from the same working directory.
A working directory is the directory from which a binary is started, it is not some arbitrary value that you can change.
I suspect these jar files where meant to be run as standalone applications and expected the filesystem resources they are trying to access to be in the same location as themselves.
Any filesystem resources would have to be moved to the location of your java.exe so that the correct file path resolution can result from your jars.
Standard Servlet project requires external libraries to be placed in the 'WEB-INF/lib' directory under project root. You can search google for 'servlet directory structure' and do your own research for more information. Shared libraries between web applications can be placed in the 'lib' directory under tomcat root, they should be picked up by tomcat jvm. My recommendation would be to keep the dependencies project specific, because you may need different versions in different projects in the future.
Update:
Read this page on the tomcat documentation, it will explain exactly how the project should be structured, and how to add a library that will be shared across all web applications:
http://tomcat.apache.org/tomcat-7.0-doc/appdev/deployment.html
Update 2:
The following Stackoverflow link explains several options how to add a static file to your web application, that will available at runtime.
https://stackoverflow.com/a/2161583/940754
Update 3:
Add a path to the classpath using the project's manifest:
http://docs.oracle.com/javase/tutorial/deployment/jar/downman.html
I have added some files that I need to be downloaded to the Application start up path. So I set Build Action as content now the files have been copied some where
C:\Documents and Settings\TestUser.ANNAM\Local Settings\Apps\2.0\Data\HVDRBMY5.8AA\858AT9VM.TNP\test..tion_2d7cfc137d9c2c74_0001.0013_432bd4561850d290\Data
How can access file from the application. My problem since it is a dynamic path will it be same folder count so that we can use like ....\Data\ Some think like this
You can use My.Application.Info.DirectoryPath this will return the directory where the application is stored.
I have added some files that I need to be downloaded to the Application start up path. So I set Build Action as content now the files have been copied some where
C:\Documents and Settings\TestUser.ANNAM\Local Settings\Apps\2.0\Data\HVDRBMY5.8AA\858AT9VM.TNP\test..tion_2d7cfc137d9c2c74_0001.0013_432bd4561850d290\Data
How can access file from the application. My problem since it is a dynamic path will it be same folder count so that we can use like ..\..\Data\ Some think like this
Application.UserAppDataPath gets the path for the application data of a user.
Application.StartupPath gives you the path for the executable file that started the application, not including the executable name.
Starting with one of these, you should be able to use System.IO to manipulate the paths until you get the folder where your data files are.