I have a java application in Eclipse that uses an eml file like so
File matches = new File("matches.xml");
The file is located in the default package as all the other classes. When I create the JAR it bundles in the XML file with it. My application require me to be able to make changes to the XML file. How can I set it up so the JAR can reference the XML file outside of itself?
My application require me to be able to make changes to the XML file.
Then you will need to extract it from the Jar and save it somewhere on the local file system.
See How can an app use files inside the JAR for read and write? for more details.
If you're using new File("matches.xml") that won't use a file within a jar file at all. It will only look on the external file system.
If you need to be able to use an external file if it's present, or the version in the jar file as a fallback, you'll need to test for the file's existence (File.exists()) and use Class.getResourceAsStream("matches.xml") for the fallback behaviour.
As you want to keep the file outside the jar and want to update it so the jar can read, so you can put the file in the same directory where the jar is and use the following code to access the file
FileInputStream file = new java.io.FileInputStream("matches.xml");
So this can be the directory structure.
- matches\
- matches.jar
- matches.xml
Related
I work on a Java console application. There is a property in my application.properties file, which contains another file name as a value of a property, like
my.file.location=file:myDir/myFileName
In the code I try to get the file like this:
#Value("${my.file.location}")
private File myfileLocation;
If I start the application from the directory, which contains jar file, the file is resolved, but when I run my application from a different location, the file location is not valid.
I can't have this file on classpath, it must be external to the jar file.
How can I make the file path to be relative to my jar file and not to the current working directory?
I believe this has nothing to do with Spring right? You just want to load configuration file, that is inside your application, unpacked, so the user can modify it, ok?
First, you may try to always setup the working directory, which I believe is more "standard" solution. In windows you can make a link, that specifies the Start in section and contains the path to your jar file (or bat or cmd, whatever).
If you insist on using the jar path, you could use How to get the path of a running JAR file solution. Note, that the jar must be loaded from filesystem:
URI path = MySpringBean.class.getProtectionDomain().getCodeSource().getLocation().toURI();
File myfileLocation = new File(new File(path).getParent(), "/myDir/jdbc.properties");
How can I access the content of the jar file which has been started. I want to create a big jar file which contains everything I need and then during runtime I want to copy some files of my jar into an external folder. Is this possible?
You want this.getClass().getClassLoader().getResourceAsStream. Example:
InputStream config =
this.getClass().getClassLoader().getResourceAsStream("config.txt");
The files in the JAR are not accessible as files, so you must use getResourceAsStream to read them. See access files and folders in executable jars how to access the files within the jar.
Following, you use the inputstream to write the files onto the file system.
See:
Easy way to write contents of a Java InputStream to an OutputStream
http://www.mkyong.com/java/how-to-convert-inputstream-to-file-in-java/
You can access any file via the ClassPath using the classloader. Start with Class.getResourceAsStream
I am trying to use a jar file which itself is a web application in another web project. In my jar which i have created using eclipse's export to jar functionality, I have stored a csv file in a folder. To use relative paths in the code in the jar I access it using
MyClass.class.getResource(ApplicationConstants.ALIASESFILE).getPath();
and this works fine when I deploy (glassfish) and use the project as a separate application. But when I am using the same from within another project, it gives a path as shown below
D:\javaProjects\AutomodeGS_Prachi\lib\internal\RESTWSGS.jar!\aliases\aliases.csv
I am getting a file notfound exception.What could be wrong?
The getResource() method is returning a "jar:" URL. The path component of that URL is not a normal filesystem pathname, and can't be opened directly using Java's file classes.
The simple way to do this is to use Class.getResourceAsStream(...) to open the stream. If you need an "identifier" for the JAR entry, use Class.getResource(...), but then open the stream using URL.openStream().
This works fine from glassfish may be because glassfish has exploded jar on file system so that your csv file is acutually a file to the file system,
if you try to read it from another project it fails because the jar containing your file is in classpath that is fine, but the csv file is under jar file and it is no longer a File
You can read it as Stream
InputStream is = MyClass.class.getResourceAsStream(ApplicationConstants.ALIASESFILE);
Im trying to get the file path of a document that is packaged as a resource in a jar file so that i can display it in a swing application. The way I have it now works when I run it from eclipse but if I export it to a runnable jar file I can't access the the documents that are packaged in the jar file. How can I get the file path of the document when its inside the jar file?
Here is the line of code showing how I am trying to access the document:
File document = new File(getClass().getResource("/resources/documents/document.pdf").getPath());
The only kind of "file path" that exists for something inside a JAR file is the path relative to the root of the JAR. But in your case it seems that you know it already (it's "/resources/documents/document.pdf"). Files inside a JAR file have no path that you can use to access them directly as they don't exist within the real file system. You need to use either getResource() or getResourceAsStream() to access them. I don't remember right now which classes are used for images in Swing, but look closely at those classes - they should have overloaded methods that accept something like InputStream or URL instead of file path.
I'm working with text files on Java. On Ubuntu 10.
But, I'm having problems with path dir.
Example:
saveFile("textFile.txt","abc");
This abstract function basically put "abc" on "textFile.txt".
I compile this file, and create a jar file (using NetBeans).
When I run the app, and call saveFile("textFile.txt","abc"), textFile.txt is saved on \home. I don't want this. I want that textFile.txtgo to pathDir inside jar file.
How do I write in this file, this same way?
When reading resources from a JAR file, you cannot use the File API. Instead, you use Class.getResourceAsStream(), like this:
reader = new InputStreamReader(MyClass.class.getResourceAsStream(
"/apathdir/textFile.txt"), "UTF-8");
Note also how the encoding is specified. FileReader does not allow that, which is why it should usually be avoided.
Iwant to know, if fileName =
"textFile.txt", what is the path dir
of this file?
If you only use a bare file name (without giving a directory), the JVM will look for the file in the current directory of the JVM process; that is usually the directory you ran the JVM (the java executable) from.
how do i do to set
/apathdir/textFile.txt?. apathdir is a
directory that is inside jar file.
I tried: fileName = "/apathdir/textFile.txt", but doesn't works.
If you want to load a file from inside a JAR file, you cannot load it using FileReader. You need to use ClassLoader.getSystemResourceAsStream() (or Class.getResourceAsStream). See e.g. this article for an explanation:
http://www.devx.com/tips/Tip/5697