I need to read an xml file which is outside the jar.I don't want to use relative path. It's an absolute path on unix box. As far as I understand, I guess there are 2 ways:-
1) Change the manifest file of jar and add that location in the classpath. In this case can I use getResource()?
2) Hardcode the path in getResourceAsStream(). My path is going to be something like this "/usr/local/folder1/folder2/".I read that I can't use getResource() for anything outside classpath and I need to use getResourceAsStream().
Which of these ways is a better approach?
Yes, both getResourceAsStream and getResource are dependent on the paths visible to the class loader that loads your class or the bootstrap class. If you want to use absolute paths, just use FileInputStream.
InputStream inputStream = new FileInputStream("/usr/local/folder1/folder2/myxml.xml");
Related
I have written a code that is packed to 1.jar
with this code:
return isProd? "/etc/waze/automation/devices.json":
DeviceRepositoryFromJsonFile.class.getClassLoader().getResource("devices.json").getPath().toString();
devices.json is here:
I have another project that depends on 1.jar
however the classLoader doesn't find the local devices.json file but rather one packed in the jar
anyhow it shows the file doesn't exist.
How can I fix this? just use a absolute path anyhow?
If as in your screenshot the devices.json locate in the src/main/resources and the package have successfully treat that as the package path and put in the jar file root directory, then you can just find the file via:
DeviceRepositoryFromJsonFile.class.getResource("/devices.json");
Note the "/" slash is important to indicate that to search from the root of the classpath.
It does not answer your question directly, but it may solve your problems faster.
As far as I can see you try to detect the absolute path to json file and pass it to another method so this file could be processed. Instead, it could be done simpler:
public byte[] getDevicesJsonBytes() {
return isProd
? IOUtils.toByteArray(ABSOLUTE_PATH_TO_PROD_FILE)
: IOUtils.toByteArray(DeviceRepositoryFromJsonFile.class.getResourceAsStream(RESOURCE_CLASSPATH);
}
The common way to read classpath resources it to use getResourceAsStream on class or classLoader instance. Also, many frameworks have their own resources abstractions, but I guess you don't need them now.
I'm trying to have my application load a resource (binary file) transparently:
If the file exists under the current directory, open it.
If not, try looking in the current JAR file if applicable.
If not, try looking in other JAR files. (This is optional and I don't mind explicitly specifying which JAR files.)
So far I know of File which opens a local file and ClassLoader which has getResource* for JAR contents.
Is there a class which combines the two? If not, how should I go about writing it myself? Should I write a ClassLoader which also checks the local filesystem? Using File? (I'm very unfamiliar with Java and don't even know what's a good type to return. InputStream?)
Thanks
P.S. By "file" I mean "path", e.g. "data/texture1.png".
Doing #1 and #3 is pretty easy. Doing #2 (just looking in the current JAR only) is much harder as it requires you figuring out what JAR you
If you wanted to check the filesystem first, otherwise load from classpath, it would be something like:
public java.io.InputStream loadByName(String name) {
java.io.File f = new java.io.File(name);
if (f.isFile()) {
return new FileInputStream(f);
} else {
return getClass().getResource(name);
}
}
If you want to prefer loading from the same JAR file first, you will need to figure out where it is. Check out Determine which JAR file a class is from for more info on figuring out the JAR file you want to load the resource from.
A URLClassLoader should be able to load both and try the file path first if the file path is on the class path ahead of the jar.
Regarding your comments:
I know that relative jar URLs don't
work. That's why the Spring guys came
up with the Resource abstraction.
Read about it here.
You might want to check the answers
to this Question: Loading a file
relative to the executing jar
file. The problem is similar to
yours.
Current jar file and current directory are not concepts in the JVM like they are when you're running a shell script. You would need to specify a directory to be used for loading the files that you're interested in, such as with a system property while executing the JVM:
java -Ddirectory.to.scan=/home/aib
Then retrieve this property:
String dir = System.getProperty("directory.to.scan");
Now when talking about JAR files, all JAR files specified explicitly on the classpath when you start the JVM are loaded by the ClassLoader. You can get the ClassLoader of a specific class by:
InputStream is = <Your class>.class.getClassLoader().getResourceAsStream("binary file");
Note that any jar file loaded by the current class loader is searched.
Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.
ClassLoader, Absolute File Paths etc etc
Suppose I have a project directory structure as,
MyProject--
--dist
--lib
--src
--test
I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.
So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"
Suppose I do this
java.io.File file = new java.io.File(""); //Dummy file
String abspath=file.getAbsolutePath();
I get the current working directory which is not necessarily project root.
Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.
Hope the problem is clear.
You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.
Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.
To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.
The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.
Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.
That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");
If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");
First, make sure the lib directory is in your classpath. You can do this by adding the command line parameter in your startup script:
$JAVA_HOME/bin/java -classpath .:lib com.example.MyMainClass
save this as MyProject/start.sh or any os dependent script.
Then you can access the textfile.txt (as rightly mentioned by Mark) as:
// if you want this as a File
URL res = getClass().getClassLoader().getResource("text/textfile.txt");
File f = new File(res.getFile());
// As InputStream
InputStream in = getClass().getClassLoader()
.getResourceAsStream("text/textfile.txt");
#Mark is correct. That is by far the simplest and most robust approach.
However, if you really have to have a File, then your best bet is to try the following:
turn the contents of the System property "java.class.path" into a list of pathnames,
identify the JAR pathname in the list based on its filename,
figure out what "../.." is relative to the JAR pathname to give you the "project" directory, and
build your target path relative to the project directory.
Another alternative is to embed the project directory name in a wrapper script and set it as a system property using a -D option. It is also possible to have a wrapper script figure out its own absolute pathname; e.g. using whence.
My build path in Eclipse looks like this:
ProjectName
-- WEB-INF
-- classes
-- myClass.class
-- configs
-- myConfig.xml
My absolute path to the config currently looks like this:
C:\Development\Java\ProjectName\WEB-INF\configs\myConfig.xml
I'm using JAXB for the binding, and it is expecting a FileInputStream. The FileInputStream needs to be a stream for the XML config file. However, I can't figure out how to get the FileInputStream for my config, and I keep getting a FileNotFoundException.
I want this config to be loaded in such a way that someone doesn't have to hardcode the path to the config because I plan on releasing the project open source. I see a lot of examples where someone just hardcodes the full absolute path, but I need it to be something more flexible "like" this:
new FileInputStream("/WEB-INF/configs/myConfig.xml");
Thanks!
I'd recommend putting that myConfig.xml in the WEB-INF/classes directory instead and loading it via the class loader, since it's in the classpath. Calling getResourceAsStream() on the servlet context will return an InputStream that you can use. It's relative to the context root, so you can pick up that WAR and put it anywhere - your code will still work.
You can get the input stream of relative paths under WEB-INF fairly easily with ServletContext.getResourceAsStream() method (javadoc here) and its variants.
For example, in your selvlet code you can do this:
getServletContext().getResourceAsStream("/WEB-INF/configs/myConfig.xml") // your file
getServletContext().getResource("/WEB-INF") // URL to WEB-INF dir
Is it possible to add a relative directory (ie, foo/bar/plugh) to the java classpath and use
InputStream in = getClassLoader().getResourceAsStream("xyzzy.properties");
To fetch foo/bar/plugh/xyzzy.properties?
My classpath looks like this:
foo.jar;foo/bar/plugh;xyz.jar
And I am able to use classes and resources from both foo and xyz jars but not from the plugh directory. In those cases, in is always null.
I can't get this to work and am not sure if this is just unsupported, I am missing something subtle or if I'm doing something wrong. Do I need to use an absolute path?
Maybe I'm misunderstanding what you're trying to do, but if you have a folder in your classpath, that means all the files underneath it should be in the classpath as well. If not, you can always pass each .properties file on the class path.
But either way, since the file/folder that contains the file is in the classpath, you should just be able to do:
InputStream in = new FileInputStream("classpath:xyz.properties")
And since "foo/bar/plugh" is in the classpath, one of the places it will look for xyz.properties is in "foo/bar/plugh".