I feel like this is a very stupid question, but i've looked for 20 minutes now and i can't for the life of me figure it out, since a lot of online solutions to it seem to not be supported anymore.
How do i get a File object of a file that is in a different package?
I need it to be a File object because that's what a need to pass on to another method.
I want to test a method that gets an argument of the type File. The file i want to use to test it is in a different package: puu.sh/8pJAg/b177243091.png (testdata). I can't seem to access it, i keep getting a filenotfoundexception when i do new File("testdata/A000100")
Packages don't contain files, they contain classes and resources. This is because a package doesn't necessarily exist on the filesystem as a directory (it could be part of a .jar file, or it could be downloaded from a network location via any of several protocols, or it could be generated on the fly by a custom classloader -- Java doesn't actually care). Your options seem to be:
Change the method to accept a stream, and use Class.getResourceAsStream(),
Copy the stream to a temporary file, call your method, then delete the temporary file once it is finished, or
Stop storing the data in a package, and make a directory for it somewhere where you can guarantee it will be present.
Related
I have a java multiple modules sbt project and some of them contains a resources folder.
module-1
resources
template.xls
module-2
resources
other.xls
After packaging I got :
lib/module-1.jar
lib/module-2.jar
And I run my program like any other java application : java -cp "lib/*" MainClass.
My problem is accessing the template.xls from module-2.jar.
At first, I've tried the lines below to get my template :
URI template = getClass().getResource("/template.xls").toURI();
Files.newInputStream(Paths.get(template), StandardOpenOption.READ);
In development mode it works. But not on the server (after deployment), it cannot find the resource.
java.nio.file.FileSystemNotFoundException: null
[jar:file:/.../lib/module-1.jar!/template.xls]
After some research I modified my accessing code like the following to get it works in both modes (development and deployed) :
InputStream templateIS = getClass().getResourceAsStream("/template.xls");
I cannot understand why !
What is the difference between the two methods ?
Files.newInputStream, as the name suggests, can open files. It cannot open anything else. It's just for files.
The concept of an InputStream is much more abstract. When you open a file for reading, you get an inputstream, yup. But you also get inputstreams for many other things: Reading from a network connection; reading the unpacked contents of zip files. Reading the output of a decryption operation. The name says it all really: It's input, and it's a stream of data. Which applies to so much more than 'file on a filesystem'.
Paths.get(template) produces a path object, which represents a file on the file system. If template is derived from a URI, this does not work unless the URI that you have so happens to be a URI to a file object; most URIs are not to file objects.
Putting it all together, in your first sample, you find a resource on the classpath (which can be files, but don't have to be. For example, they could be entries in a jar file), you then ask its URI, feed it to the Paths API to turn it into a Path object, and then ask the Files API to turn this into an InputStream, which only works if the URI represents a file.
In the second snippet, you just ask the classloader system to get you an inputstream. It knows how to do that (after all, java has to load those class files!). If the resource you're asking for so happens to be represented by a file, it's going to do, internally, more or less the same thing as your first snippet: Use the Files API to open the file for reading. But if it's anything else, it knows how to do that too – it also knows how to get resources across a network, from inside jar files, generated on-the-fly – the concept of class loading (which is what class.getResource lets you access) is abstracted away.
NB: You're using it wrong. The proper way is ClassYouAreWritingIn.class.getResource and ClassYouAreWritingIn.class.getResourceAsStream; getClass().getResource is not correct; that breaks when subclassing, whereas the correct form doesn't.
From docs,
The method getResource() returns a URL for the resource. The URL (and
its representation) is specific to the implementation and the JVM
(that is, the URL obtained in one runtime instance may not work in
another). Its protocol is usually specific to the ClassLoader loading
the resource. If the resource does not exist or is not visible due to
security considerations, the methods return null.
If the client code wants to read the contents of the resource as an
InputStream, it can apply the openStream() method on the URL. This is
common enough to justify adding getResourceAsStream() to Class and
ClassLoader. getResourceAsStream() the same as calling
getResource().openStream(), except that getResourceAsStream() catches
IO exceptions returns a null InputStream.
So, getResourceAsStream() the same as calling getResource().openStream(), except that getResourceAsStream() catches IO exceptions returns a null InputStream.
Your first scenario doesn't work because the template.xls is packaged within a jar file. It's not a file on the filesystem (whereas it likely is in your development env prior to packaging as a file). As such the Files API can't find it.
Class.getResourceAsStream() makes use of the classloading mechanism, and the class (obviously) is loaded from the .jar file.
I need to store data into files inside .jar file and read it again.
I know that I can use Class.getResourceAsStream() method but it returns an InputStream that I can read from. But I look for a way to write.
I need to store data into files inside .jar file and read it again
No you don't.
Instead store the 'default' file inside the Jar. If it is changed, store the altered file in another place. One common place is a sub-directory of user.home. When checking for the file, first check the existence of an altered file on the file system, and if it does not exist, load the default file.
Note that it is generally better to describe the goal, rather than the strategy. 'Store changed file in Jar' is a strategy, whereas 'Save preferences between runs' might be the goal.
Related: What is the XY problem?
This is not really supported. In principle, you could operate on the jar file, but there's no guarantee the new contents would be correctly loaded. Let your build tools manage the jar file -- and choose something else for persistent storage managed by your program itself. Like a file.
It is unlikely that you can change a loaded jar safely. Changing it while it is being used is not a good idea.
If you want to do this, use a plain file system directory and add/remove files from it. Even this may not work as you expect.
You can manipulate any jar file using the package java.util.jar (or indeed just java.util.zip). As files inside a jar will be compressed, this isn't the most time efficient way for you to store data.
You should probably use a directory somewhere else (e.g. System.getProperty("user.home") + "/.myProgram") or see java.util.prefs.
Class.getResource() returns a URL. Theoretically, you can use this URL to create your InputStream and OutputStream. But in most cases, the generated JAR is a read-only file (or archive). So your application might trip when trying to use it.
I need to store data into files inside .jar file and read it again.
I know that I can use Class.getResourceAsStream() method but it returns an InputStream that I can read from. But I look for a way to write.
I need to store data into files inside .jar file and read it again
No you don't.
Instead store the 'default' file inside the Jar. If it is changed, store the altered file in another place. One common place is a sub-directory of user.home. When checking for the file, first check the existence of an altered file on the file system, and if it does not exist, load the default file.
Note that it is generally better to describe the goal, rather than the strategy. 'Store changed file in Jar' is a strategy, whereas 'Save preferences between runs' might be the goal.
Related: What is the XY problem?
This is not really supported. In principle, you could operate on the jar file, but there's no guarantee the new contents would be correctly loaded. Let your build tools manage the jar file -- and choose something else for persistent storage managed by your program itself. Like a file.
It is unlikely that you can change a loaded jar safely. Changing it while it is being used is not a good idea.
If you want to do this, use a plain file system directory and add/remove files from it. Even this may not work as you expect.
You can manipulate any jar file using the package java.util.jar (or indeed just java.util.zip). As files inside a jar will be compressed, this isn't the most time efficient way for you to store data.
You should probably use a directory somewhere else (e.g. System.getProperty("user.home") + "/.myProgram") or see java.util.prefs.
Class.getResource() returns a URL. Theoretically, you can use this URL to create your InputStream and OutputStream. But in most cases, the generated JAR is a read-only file (or archive). So your application might trip when trying to use it.
I need to store data into files inside .jar file and read it again.
I know that I can use Class.getResourceAsStream() method but it returns an InputStream that I can read from. But I look for a way to write.
I need to store data into files inside .jar file and read it again
No you don't.
Instead store the 'default' file inside the Jar. If it is changed, store the altered file in another place. One common place is a sub-directory of user.home. When checking for the file, first check the existence of an altered file on the file system, and if it does not exist, load the default file.
Note that it is generally better to describe the goal, rather than the strategy. 'Store changed file in Jar' is a strategy, whereas 'Save preferences between runs' might be the goal.
Related: What is the XY problem?
This is not really supported. In principle, you could operate on the jar file, but there's no guarantee the new contents would be correctly loaded. Let your build tools manage the jar file -- and choose something else for persistent storage managed by your program itself. Like a file.
It is unlikely that you can change a loaded jar safely. Changing it while it is being used is not a good idea.
If you want to do this, use a plain file system directory and add/remove files from it. Even this may not work as you expect.
You can manipulate any jar file using the package java.util.jar (or indeed just java.util.zip). As files inside a jar will be compressed, this isn't the most time efficient way for you to store data.
You should probably use a directory somewhere else (e.g. System.getProperty("user.home") + "/.myProgram") or see java.util.prefs.
Class.getResource() returns a URL. Theoretically, you can use this URL to create your InputStream and OutputStream. But in most cases, the generated JAR is a read-only file (or archive). So your application might trip when trying to use it.
I am doing a project in java and in that i need to add and modify my
text file at runtime,which is grouped in the jar.
I am using class.getResourceAsStream(filename) this method we
can read that file from class path.
i want to write into the same textfile.
What is the possible solution for this.
If i can't update the text file in jar what other solution is there?
Appreciate any help.
The easiest solution here is to not put the file in the jar. It sounds like you are putting files in your jar so that your user only needs to worry about one file that contains everything related to that program. This is an artificial constraint and just add headaches.
There is a simple solution that still allows you to distribute just the jar file. At start up, attempt to read the file from the file system. If you don't find it, use default values that are encoded in you program. Then when changes are made, you can write it to the file system.
In general, you can't update a file that you located using getResourceAsStream. It might be a file in a JAR/ZIP file ... and writing it would entail rewriting the entire JAR file. It might be a remote file served up by a Url classloader.
For your sanity (and good practice), you should not attempt to update files that you access via the classpath. If you need to, read the file out of the JAR file (or whatever), copy it into the regular file system, and then update the copy.
I'm not saying that it is impossible to do this in all cases. Indeed, in most normal cases you can do it with some effort. However, this is not supported, and there are no standard APIs for doing this.
Furthermore, attempts to update resources are liable to cause anomalies in the classloader. For example, I'd expect resources in JAR files to not update (from the perspective of the application) until the application restarted. But resources in exploded JAR files probably would update ... though new resources might not show up.
Finally, there are cases where updating a resource is impossible:
When the user doesn't have write access to the application's installation directory. This is typical for a properly administered UNIX / Linux machine.
When the JAR file is fetched from a remote server, you are likely not to be able to write the updates back.
When you are using an arbitrary custom classloader, you've got no way of knowing where the actual bytes of an updated resource should be stored, and no way of storing them.
All JAR rewriting techniques in Java look similar. Open the Jar file, read all of it's contents, and write a new Jar file containing the unmodified contents (and the modifications you whished to make). Such techniques are not advisable for a Jar file on the class path, much less a Jar file you're running from.
If you decide you must do it this way, Java World has a few articles:
Modifying Archives, Part 1
Modifying Archives, Part 2
A good solution that avoids the need to put your items into a Jar file is to read (if present) a properties file out of a hidden subdirectory in the user's home directory. The logic looks a bit like this:
if (the hidden directory named after my application doesn't exist) {
makeTheHiddenDirectory();
writeTheDefaultPropertiesFile();
}
Properties appProps = new Properties();
appProps.load(new FileInputStream(fileInHiddenDir));
...
... After the appProps have changed ...
...
appProps.store(new FileOutputStream(fileInHiddenDir), "Do not modify this file");
Look to java.util.Properties, and keep in mind that they have two different load and store formats (key = value based and XML based). Pick the one that suits you best.
If i can't update the text file in jar what other solution is there?
Store the information in any of:
Cookies
The server
Deploy the applet using 1.6.0_10+, launch it using JWS and use the PersistenceService to store the information. Here is my demo. of the PersistenceService.
Also, if your users will agree to a trusted applet (which seems overkill for this), you might write the information to a sub-directory of user.home.