I searched a lot but couldn't find anything. I want to validate a selected xml file with an schema file. My problem is working relative path is config folder of the Glassfish server.
I tried to get url of the schema file which I copied into same package of the class I run this code but still I can't find the file. I only able to make it work is by copying the file into server config file.
URL url = ClassLoader.getSystemResource("/schema.xsd");
File file = new File(url.getPath());
But url returns null. How can I find this file and even if I find I believe file object can not be created because my relative path is /home/user/glassfish3/glassfish/domains/domain1/config. Any ideas how to solve this problem?
Thanks for any help in advance.
The leading slash makes the path absolute. If you remove it, leave just the file name and have that file in the domain/config dir then it should work.
Note that if you'd like this file somewhere else you could also provide a relative path stepping back your directory tree e.g. ../../prj/xsd/schema.xsd
Related
i have following line
File file = ResourceUtils.getFile("classpath:calculation.csv");
and i also tried
File file = ResourceUtils.getFile("classpath:/calculation.csv");
but both will throw an error
java.io.FileNotFoundException: class path resource [calculation.csv] cannot be resolved to absolute file path because it does not exist
but i do have calculation.csv in by resources folder..
why is this?
I need to read file from resources folder, and it should also work in server enviroment
EDIT:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("calculation.csv").getFile());
works just as fine, so not at all..
EDIT2:
tried with folder.. i have both calculation.csv and csv/calculation.csv in my resources folder now..
none of the above work, with /csv/ added.
what kind of path does this thing want?!
EDIT3:
aaand
File file = new ClassPathResource("calculation.csv").getFile();
is also no go, what even is this..
Loading file (as FILE) wont work. You must use it as resource. Files inside JAR will not work as file anyway. This is also what your "check" code shows.
classLoader.getResource("calculation.csv") works, because you are using classloader to get resource, not filesystem to get file (which is what File api does). It could work, if you would deal with non packed application. Once you pack your app into JAR, file path will be like your/path/to/jar.jar!someResource - note ! mark (and that is what you would see as well). So basicly it will return File instance, you that you wont be able to use anyway, as file system has no access to it.
You could alternatively try to extract it first with ResourceUtiuls#extractJarFileURL(URL jarUrl) and then use extracted file.
I think, that in most cases Class#getResourceAsStream is the way to go and I think that it should fit your needs as well to read content of resource.
I have config.properties file in src/main/resources folder having a property as "driverFilePath=C:\\VSP\\workspace\\Drivers\\IEDriverServer_Win32_3.5.0\\IEDriverServer.exe".
Here I'm currently reading it by mentioning full path, now I want to read only through relative path as "Drivers\\IEDriverServer_Win32_3.5.0\\IEDriverServer.exe" instead, as the drivers are also located in svn repo.
I'm using the code as
File file = new File(driverFilePath); to get the path from local.Please suggest a way so that I can mention only relative path like "Drivers/IEDriverServer_Win32_3.5.0/IEDriverServer.exe" which will match both my local and also the svn location.Hence I won't be making any change in my code.
I am using IDEA 14 to follow a simple Java tutorial (JDBC). As part of this, I am storing some configuration properties in a file called jdbcTutorial.properties. When I put this in the root directory of the project, I can read it with the following:
Properties props = new Properties();
props.load(new FileInputStream("jdbcTutorial.properties"));
However, as soon as I move it to any other directory in the project, I get the error "No such file or directory". This happens regardless of whether I specify a relative or absolute path:
Maybe there are more standard ways to use config files, but I would really like to understand the behavior I am observing. Thanks for helping!
By default root directory would be added to your project's build path. Since the directory in which you are putting the file is not added in your project's build path jvm is unable to find the file. You have two options:
Add the folder where you are putting your prop to build path.
Access the file with full path i.e. /home/user/workspace/....
When you build a project, IDEA takes the files in the resources directory and puts them in the executable jar. So to get an input stream from that file, you need to get it directly from inside the jar. Instead of FileInputStream, use
getClass().getResourceAsStream("jdbcTutorial.properties")
When no path is supplied for a file, java assumes that this file is in the project's root folder. So the relative path "." always points to this folder. When the file is somewhere else, put the appropriate relative path before the path name, like "./files/configuration/jdbcTutorial.properties".
It work also when you put an absolute file path before the path name, like "C:/Users/Me/Documents/Java/workspace/thisProject/files/configuration/jdbcTutorial.properties".
Problem statement:
I have a jar file with a set of configuration files in a package mycompany/configuration/file/.
I don't know the file names.
My intention is to load the file names at runtime from the jar file in which they are packaged and then use it in my application.
As far as I understood:
When I use the ClassLoader.getResources("mycompany/configuration/file/") I should be getting all the configuration files as URLs.
However, this is what is happening:
I get one URL object with URL like jar:file:/C:/myJarName.jar!mycompany/configuration/file/
Could you please let me know what I am doing wrong ?
For what you are trying to do I don't think it is possible.
getResource and getResources are about finding named resources within the classloader's classpath, not listing values beneath a directory or folder.
So for example ClassLoader.getResources("mycompany/configuration/file/Config.cfg") would return all the files named Config.cfg that existed in the mycompany/configuration/file path within the class loader's class path (I find this especially useful for loading version information personally).
In your case I think you might almost have half a solution. The URL you are getting back contains the source jar file (jar:file:/C:/myJarName.jar). You could use this information to crack open the jar file a read a listing of the entries, filtering those entries whose name starts with "mycompany/configuration/file/".
From there, you could then fall back on the getResource method to load a reference to each one (now that you have the name and path)
On my vps, I want to upload a file to the Logos directory.
The directory structure is as follows on my vps -
/home/webadmin/domain.com/html/Logos
When a file is uploaded through my jsp page, that file is renamed, and then I want to put it into the Logos directory.... but I can't seem to get the path right in my servlet code.
Snippet of servlet code -
String upload_directory="/Logos/"; // path to the upload folder
File savedFile = new File(upload_directory,BusinessName+"_Logo."+fileExtension);
//.....
//file saved to directory
//.....
I've tried many variations, but still fail. What is the proper way to specify the path?
Edited
The problem with using getServletContext() is that it returns the path to the directory where Tomcat and my webapp is...whereas I want to reach the directory where my html and image files are - under the root directory of the vps. How do I specify that path?
String server_path = getServletContext().getRealPath("/"); // get server path.
//server_path = /opt/tomcat6/webapps/domain.com/
String upload_directory = "Logos/"; // get path to the upload folder.
String complete_path = server_path + upload_directory; // get the complete path to the upload folder.
//complete_path = /opt/tomcat6/webapps/domain.com/Logos/
File savedFile = new File(complete_path,"NewLogo.jpg");
//savedFile = /opt/tomcat6/webapps/domain.com/Logos/NewLogo.jpg
It's a common practice to make the path for storage configurable - either via some application.properties file, or if you don't have such a properties file - as a context-param in web.xml. There you configure the path to be the absolute path, like:
configuredUploadDir=/home/webadmin/domain.com/html/Logos
Obtain that value in your code (depending on how you stored it), and have:
File uploadDir = new File(configuredUploadDir);
Note: make sure you have the permissions to read and write the target directory.
You can use following code in any jsp or servlet.
1) String serverPath= getServletContext().getRealPath("/");
This will give you full path of the server from root directory to your web application directory.
For me its: "D:\local\tomcat-6.0.29\webapps\myapp" when I sysout from myapp application.
Once you got the whole real path for the server system as above you can get the path relative to your directory. So if I have some data file in myapp\data - I can get it appending \data\filename to the serverPath which we got earlier.
This will work in all situation even you have multiple servers installed on the same system.
2) You can get server home from system properties using
System.getProperty("TOMCAT_HOME")
and then can use this absolute path in your program
3) To pass absolute directory path to any servlet using <init-param>
Hope this will work for you.
Well, the problem is: the File constructor doesn't create the file, only prepares to for the creation, then, after you construct a file instance you must invoke the method createNewFile(), and thats all.
The path "/Logos/" will attempt to create the file in the root of your system, which is not what you want. Look at the ServletContext.getRealPath() method.