i know this question has been asked several times, but i think my problem differs a bit from the others:
String resourcePath = "/Path/To/Resource.jar";
File newFile = new File(resourcePath);
InputStream in1 = this.getClass().getResourceAsStream(resourcePath);
InputStream in2 = this.getClass().getClassLoader().getResourceAsStream(resourcePath);
The File-Object newFile is completely fine (the .jar file has been found and you can get its meta-data like newFile.length() etc)
On the other hand the InputStream always return null.
I know the javadoc says that the getResourceAsStream() is null if there is no resource found with this name, but the File is there! (obviously, because it's in the File-Object)
Anyone know why this happens and how i can fix it so that i can get the .jar File in the InputStream?
The getResourceAsStream() method doesn't load a file from the file system; it loads a resource from the classpath. You can use it to load, for example, a property file that's packaged inside your JAR. You cannot use it to load a file from the file system.
So, if your file resides on the file system, rather than in your JAR file, better use the FileInputStream class.
Related
I'm just trying to read in a simple .txt file into my java project using this.class.getResourceAsStream(filename). I have several files within main/resources, and almost all of them return an object when I try to get them as an input stream. The only object I can't read in is my text file.
I have placed the file with all of the other resource files that are readable by the classloader, but it appears this file wasn't placed in the class' classLoader for whatever reason. If I unzip the jar, the file is still included with the jar in the same directory as all of the other resources, so it seems to be being built correctly.
I guess what I'm asking is at what point do I tell Java what files I want to be included as a resource in a class' ClassLoader? Is it something that should be done when the jar is built if things are in the correct place (i.e main/resources)?
Here is what the code looks like, and it's respective return values, when running for the file it can find and the file it can't, that are both located in the same place.
// This is not found. Both are placed at src/main/resources
def tmpDict = this.class.getResourceAsStream("dict.txt")
println tmpDict // null
// This is found
def tmpDict2 = this.class.getResourceAsStream("calc.config")
println tmpDict2 // sun.net.www.protocol.jar.JarURLConnection$JarURLInputStream#2dae5a79
Without more info i'd say the path is wrong. when i used just "file.txt" for the path it got NPE
i used this method to read from the stream. The file was located at \src\main\resources\static\file.txt
This worked in eclipse, packaged into jar and worked there too.
public String getFile() throws Exception {
InputStream in = Controller.class.getClassLoader().getResourceAsStream("static/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in, Charset.defaultCharset()));
StringBuilder out = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
out.append(line);
}
return out.toString();
}
Some very Basic checks:
The file must be case-sensitive (as it is not a Windows file), and special characters of the file name might be cumbersome. Also check that the file in your project has the file extension not twice (.txt.txt - Windows hiding the extension).
Check that getResourceAsStream("/a/b/c/A.txt") indeed gives a null.
If not the reading might go wrong on the encoding.
This is a chunk of data I'd like to access by a method.
I'm doing the following to read my file:
String fileName = "file.txt"
InputStream inputStream = new FileInputStream(fileName);
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
My file.txt is in the same package, but I still get FileNotFoundException.
I didn't use a path url to point to the file because I thought since this it going to be an android application, hard-coding the path might not work when deployed... Please correct me if I am wrong. Thanks bunch!
This shows how to do that. https://stackoverflow.com/a/14377185/2801237
Also the 'package' your class is in has nothing to do with the 'path' where the file is being executed from. (two different concepts, 'package' = folder hierarchy of java source code files), 'path' = location on a filesystem of a specific file, your APK is being 'executed' in a particular place, and the location it writes a file is associated with that (I actually don't know where 'offhand' it writes by default, because I always get cache dir, or sd card root, etc.)
You may use:
InputStream inputStream = this.getClass().getResourceAsStream(fileName);
In my Maven project, I have a xls file in src/main/resources.
When I read it like this:
InputStream in = new
FileInputStream("src/main/resources/WBU_template.xls");
everything is ok.
However I want to read it as InputStream with getResourceAsStream. When I do this, with or without the slash I always get a NPE.
private static final String TEMPLATEFILE = "/WBU_template.xls";
InputStream in = this.getClass.getResourceAsStream(TEMPLATEFILE);
No matter if the slash is there or not, or if I make use of the getClassLoader() method, I still get a NullPointer.
I also have tried this :
URL u = this.getClass().getResource(TEMPLATEFILE);
System.out.println(u.getPath());
the console says.../target/classes/WBU_template.xls
and then get my NullPointer.
What am I doing wrong ?
FileInputStream will load a the file path you pass to the constructor as relative from the working directory of the Java process.
getResourceAsStream() will load a file path relative from your application's classpath.
When you use .getClass().getResource(fileName) it considers the location of the fileName is the same location of the of the calling class.
When you use .getClass().getClassLoader().getResource(fileName)
it considers the location of the fileName is the root - in other words bin folder.
The file should be located in src/main/resources when loading using Class loader
In short, you have to use .getClass().getClassLoader().getResource(fileName) to load the file in your case.
I usually load files from WEB-INF like this
session.getServletContext().getResourceAsStream("/WEB-INF/WBU_template.xls")
Okay, so this is the line that's returning null. What am I doing wrong while creating this FileInputStream?
FileInputStream fin = new FileInputStream(new File(getClass().getResource("data/levellocks.lv").toURI()));
The only thing that can be null there is getResource("data/levellocks.lv") which is calling the toURI call to fail
Either getClass or getResource could return null. Everything else should succeed or throw an exception.
Unless you really need a file input stream, you line can be simplified to:
InputStream in = getClass().getResourceAsStream("data/levellocks.lv");
Class.getResource() and Class.getResourceAsStream are relative to the package. To get the file relative to the root of the classpath, you can call those methods on the classloader:
InputStream in = getClass().getClassLoader().getResourceAsStream("data/levellocks.lv");
Did you make sure the file is in your binary folder, next to the .class files? Not just in your source folder next to the .java files?
I actually just dealt with this issue (I'm no expert) but try debugging and see where the constructor is trying to resolve the name to. For me, it was the package of the class. So when I put the file in the expected folder, it found it.
Would probably be different for you, as I'm using maven. But I put it in src/main/resources and it couldn't find it. When I put a folder structure in src/main/resources of com.work.hin.terminology.match (which was the package of the class), it found it.
I'm using Spring's Resource abstraction to work with resources (files) in the filesystem. One of the resources is a file inside a JAR file. According to the following code, it appears the reference is valid
ResourcePatternResolver resourceResolver = new PathMatchingResourcePatternResolver();
// The path to the resource from the root of the JAR file
Resource fileInJar = resourcePatternResolver.getResources("/META-INF/foo/file.txt");
templateResource.exists(); // returns true
templateResource.isReadable(); // returns true
At this point, all is well, but then when I try to convert the Resource to a File
templateResource.getFile();
I get the exception
java.io.FileNotFoundException: class path resource [META-INF/foo/file.txt] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/D:/m2repo/uic-3.2.6-0.jar!/META-INF/foo/file.txt
at org.springframework.util.ResourceUtils.getFile(ResourceUtils.java:198)
at org.springframework.core.io.ClassPathResource.getFile(ClassPathResource.java:174)
What is the correct way to get a File reference to a Resource that exists inside a JAR file?
What is the correct way to get a File
reference to a Resource that exists
inside a JAR file?
The correct way is not doing that at all because it's impossible. A File represents an actual file on a file system, which a JAR entry is not, unless you have a special file system for that.
If you just need the data, use getInputStream(). If you have to satisfy an API that demands a File object, then I'm afraid the only thing you can do is to create a temp file and copy the data from the input stream to it.
If you want to read it, just call resource.getInputStream()
The exception message is pretty clear - the file does not reside on the file-system, so you can't have a File instance. Besides - what will do do with that File, apart from reading its content?
A quick look at the link you provided for Resource documentation, says the following:
Throws: IOException if the resource cannot be resolved as absolute file path,
i.e. if the resource is not available in a file system
Maybe the text file is inside a jar? In that case you will have to use getInputStream() to read its contents.
Just adding an example to the answers here. If you need a File (and not just the contents of it) from within your JAR, you need to create a temporary file from the resource first. (The below is written in Groovy):
InputStream inputStream = resourceLoader.getResource('/META-INF/foo/file.txt').inputStream
File tempFile = new File('file.txt')
OutputStream outputStream = new FileOutputStream(tempFile)
try {
IOUtils.copy(inputStream, outputStream)
} catch (IOException e) {
// Handle exception
} finally {
outputStream.close()
}