I have a jar file which have 15 classes. I have to call some methods of this jar from web application. Some of the methods read .plist file.
In the jar code I have coded in such a way that all the plist files are readed once and uses multiple times(I have used some static map which hold all the properties of .plist files), but this is not happening.
Every time there is a call it is reading all .plist files again. And because of this there is "Too Many Open File Exception."
Is there any way to stop the file reading again and again. Please help me.
Can you post your code or stack trace. it looks like a logical problem . If you want to read file once just have a static flag and after reading it set it true and before reading it just check....If this is not the desired answer just let me know .
Related
Summary:
I have a program I want to ship as a single jar file.
It depends on three big resource files (700MB each) in a binary format. The file content can easily be accessed via indexing, my parser therefore reads these files as RandomAccessFile-objects.
So my goal is to access resource files from a jar via File objects.
My problem:
When accessing the resource files from my file system, there is no issue, but I aim to pack them into the jar file of the program, so the user does not need to handle these files themselves.
The only way I found so far to access a file packed in a jar is via InputStream (generated by class.getResourceAsStream()), which is totally useless for my application as it would be much too slow reading these files from start to end instead of using RandomAccessFile.
Copying the file content into a file, reading it and deleting it in runtime is no option eigher for the same reason.
Can someone confirm that there is no way to achieve my goal or provide a solution (or a hint so I can work it out myself)?
What I found so far:
I found this answer and if I understand the answer it says that there is no way to solve my problem:
Resources in a .jar file are not files in the sense that the OS can access them directly via normal file access APIs.
And since java.io.File represents exactly that kind of file (i.e. a thing that looks like a file to the OS), it can't be used to refer to anything in a .jar file.
A possible workaround is to extract the resource to a temporary file and refer to that with a File.
I think I can follow the reasoning behind it, but it is over eight years old now and while I am not very educated when it comes to file systems and archives, I know that the Java language has evolved quite much since then, so maybe there is hope? :)
Probably useless background information:
The files are genomes in the 2bit format and I use the TwoBitParser from biojava via the wrapper class TwoBitFacade?. The Javadocs can be found here and here.
Resources are not files, and they live in a JAR file, which is not a random access medium.
I want to change the automatically saved filename of a HDFS block on a cluster. So when I upload a file to the HDFS and it'll divided into blocks, they get generated names with several numbers.
I searched the code but couldn't find it.
It should be somewhere in the java code, not a configuration file, right?
Can someone may tell me in which classes this creation or definition of the filename happens?
Thanks a lot!
edit:
Ok found it.
FSNameSystem.randBlockId.nextLong()
Ok found it.
FSNameSystem.randBlockId.nextLong()
In File Monitoring process if one file came and it is processed immediately it does not check
if file is open and writing something..Then how to prevent moving of file without closing the file.
Do you have control over the program that's putting the files in the directory? Put something like ".partial" on the end of the filename while the file is still being written, and then rename it to remove the ".partial" when the writing is done. If you make the Java file-monitoring program ignore files whose names end in ".partial", it'll only see files after they've been fully written out.
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.
I have this recurrent Java JAR program tasks that tries to modify a file every 60seconds.
Problem is that if user is viewing the file than Java program will not be able to modify the file. I get the typical IOException.
Anyone knows if there is a way in Java to modify a file currently in use? Or anyone knows what would be the best way to solve this problem?
I was thinking of using the File canRead(), canWrite() methods to check if file is in use. If file is in use then I'm thinking of making a backup copy of data that could not be written. Then after 60 seconds add some logic to check if backup file is empty or not. If backup file is not empty then add its contents to main file. If empty then just add new data to main file. Of course, the first thing I will always do is check if file is in use.
Thanks for all your ideas.
I was thinking of using the File
canRead(), canWrite() methods to check
if file is in use.
Not a good idea - you'll run into race conditions e.g. when your code has used those check methods, received true return values, but then the file is locked by a different application (possibly the user) just before you open it for writing.
Instead, try to get a FileLock on the file and use the "backup file" when that fails.
You can hold a lock on the file. This should guarantee you are able to write on the file.
See here on how to use the FileLock class.
If the user is viewing the file you should still be able to read it. In this case, make an exact copy of the file, and make changes to the new file.
Then after the next 60 seconds you can either:
1) Check if the file is being viewed and if not, delete it and replace it with the earlier file, then directly update this file.
2) If it is being viewed, continue making changes to the copy of the file.
EDIT: As Michael mentioned, when working with the main file, get a lock on it first.