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Closed 10 years ago.
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Loading from JAR as an InputStream?
Is it possible for me to load a jar file from an input stream (from a url connection, for example), load it into a classloader, and execute it?
Thanks for your time.
Yes; URLClassLoader is intended for this purpose. It can load classes from an array of URLs.
URL externalJar = new URL("http://example.com/app.jar");
URL localJar = new URL("C:/Documents/app.jar");
URLClassLoader cl = new URLClassLoader(URL[]{ externalJar, localJar });
Class<?> clazz = cl.loadClass("SomeClass"); // you now can load classes
If your InputStream is not based on a URL, you could instead write the stream's contents to a temporary jar file which you then load using the above approach. (You can load a jar entirely in memory, meaning no temp file is created, but this method takes considerably more effort to do right, because you will need define a custom ClassLoader).
public static URL getJarUrl(final File file) throws IOException {
return new URL("jar:" + file.toURI().toURL().toExternalForm() + "!/");
}
This properly fetches the URL of the of the Lib because Jar files do some wonky stuff when trying to get their URL.
Related
This question already has answers here:
How to solve the java.nio.file.NoSuchFileException?
(6 answers)
Closed 9 months ago.
I try to read a file from the resources and it tells me
java.io.FileNotFoundException: file:/home/simon/IdeaProjects/KTMBlockChain/build/resources/main/certificate_template.docx (No such file or directory)
Note that the file is blue and clickable. When I click on it it also opens the file so it definietly exists in the place expected.
Code:
InputStream in = null;
try {
File file = new File(this.getClass().getClassLoader().getResource("certificate_template.docx").toString());
in = new FileInputStream(file);
IXDocReport report = XDocReportRegistry.getRegistry().
loadReport(in, TemplateEngineKind.Freemarker);
Options options = Options.getTo(ConverterTypeTo.PDF).via(ConverterTypeVia.ODFDOM);
IContext ctx = report.createContext();
ctx.put("re_wo", pdfData.getReifen());
/*ctx.put("to", invoice.getTo());
ctx.put("sender", invoice.getInvoicer());
FieldsMetadata metadata = report.createFieldsMetadata();
ctx.put("r", invoice.getInvoiceRows());*/
report.convert(ctx, options, new FileOutputStream("result.pdf"));
I dont know what to do anymore...
EDIT 1: Changed code, still not working, another error code but same problem
There are
Disk files, class File;
Resource files (read-only), on the class path, possibly packed in a jar or war.
So here is you should use a resource, and Path is a generalisation of File, and all other kind of URL paths. This caused the error.
URL url = getClass().getClassLoader().getResource("certificate_template.docx");
Path path = Paths.get(url.toURI());
List<String> x = Files.readAllLines(path); // Reading a UTF-8 text.
But docx is not text, but a binary format (actually a zip format).
You either need to use a library or just handle the file as-is, like let the operating system open it.
You could use Files.readAllBytes(); reading UTF-8 will probably cause an error as the bytes are not in UTF-8 format.
After edit of question:
in = getClass().getClassLoader().getResourceAsStream("certificate_template.docx");
A lot has been discussed already here about getting a resource.
If there is already a solution - please point me to it because I couldn't find.
I have a program which uses several jars.
To one of the jars I added a properties file under main/resources folder.
I've added the following method to the jar project in order to to read it:
public void loadAppPropertiesFile() {
try {
Properties prop = new Properties();
ClassLoader loader = Thread.currentThread().getContextClassLoader();
String resourcePath = this.getClass().getClassLoader().getResource("").getPath();
InputStream stream = loader.getResourceAsStream(resourcePath + "\\entities.properties");
prop.load(stream);
String default_ssl = prop.getProperty("default_ssl");
}catch (Exception e){
}
}
The problem (?) is that resourcePath gives me a path to the target\test-clasess but under the calling application directory although the loading code exists in the jar!
This the jar content:
The jar is added to the main project by maven dependency.
How can I overcome this state and read the jar resource file?
Thanks!
I would suggest using the classloader used to load the class, not the context classloader.
Then, you have two options to get at a resource at the root of the jar file:
Use Class.getResourceAsStream, passing in an absolute path (leading /)
Use ClassLoader.getResourceAsStream, passing in a relative path (just "entities.properties")
So either of:
InputStream stream = getClass().getResourceAsStream("/entities.properties");
InputStream stream = getClass().getClassLoader().getResourceAsStream("entities.properties");
Personally I'd use the first option as it's briefer and just as clear.
Can you try this:
InputStream stream = getClass().getClassLoader().getResourceAsStream("entities.properties")
This question already has an answer here:
Any way to get a File object from a JAR
(1 answer)
Closed 5 years ago.
I'm trying to implement a button into my project which, when clicked, automatically loads a specific file. Currently there are buttons for users selecting a file from their hard disk.
So, I downloaded the specific file and inserted it into the project. When using File f = new File("demofile") or something like this
getClass().getResource("/resources/file.txt").getFile(); the code WORKS locally.
However, when the project is packaged, a FileNotFoundException is thrown.
After much research online, there are suggestions to use something like:
InputStream is = getClass().getResourceAsStream("/resources/file.txt");
However, for this project, I need the file to be referenced as a file object so that it can be passed as an argument to other functions, such as:
in = new TextFileFeaturedSequenceReader(TextFileFeaturedSequenceReader.FASTA_FORMAT, file, DiffEditFeaturedSequence.class);
Any ideas on how I can solve this, or read a stream into a file object?
Thanks!
If you absolutely must pass a File, copy your resource to a temporary file:
Path path = Files.createTempFile(null, null);
try (InputStream stream =
getClass().getResourceAsStream("/resources/file.txt")) {
Files.copy(stream, path, StandardCopyOption.REPLACE_EXISTING);
}
in = new TextFileFeaturedSequenceReader(
TextFileFeaturedSequenceReader.FASTA_FORMAT,
path.toFile(),
DiffEditFeaturedSequence.class);
// Use the TextFileFeaturedSequenceReader as needed
// ...
Files.delete(path);
This question already has answers here:
How to read text file from classpath in Java?
(17 answers)
Closed 5 years ago.
In our application, we are sending an email during registration.
For this we have got an email template stored under /usr/local/email.html
StringWriter writer = new StringWriter();
IOUtils.copy(new FileInputStream(new File("/usr/local/email.html")),writer);
message.setContent(writer.toString(), "text/html");
Transport.send(message);
The above is working fine .
We don't want to hardcode this path (/usr/local/email.html), wanted to keep under classpath
If I keep it under classpath how can I read it?
StringWriter writer = new StringWriter();
ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("email.html");
IOUtils.copy(is,writer);
message.setContent(writer.toString(), "text/html");
Transport.send(message);
With the above code, I'm getting a java.lang.NullPointerException at
IOUtils.copy(is,writer);
If you use getClass().getResourceAsStream with a relative path, the resource is found if it's in the same package as the class.
Alternatively you can use an absolute name like "/some/package/email.html"
Why dont you use
this.getClass().getClassLoader().getResourceAsStream("email.html");
Your resource is in a fixed, known location on your classpath, so it's best to use the Class#getResourceAsStream method. You call it on a Class object, and it automatically uses the class's ClassLoader and looks for the resource in the class's package.
For example, if you have a class called com.example.myapp.Emailer and you write:
InputStream is = Emailer.class.getResourceAsStream("email.html");
it will look for com/example/myapp/email.html via the same classloader that's used to load the Emailer class. You'd put email.html in your JAR, in the same place as the compiled Emailer.class file.
I'd specifically recommend not using Thread.currentThread().getContextClassLoader() for this, because the result can vary from thread to thread, and it could potentially be a classloader that isn't able to access the JAR file that holds your resource. You want to make sure you're always using the classloader that actually loads your application's classes, and you accomplish that by calling getResourceAsStream on the Class object of one of your application's classes.
This question already has answers here:
How to find the working folder of a servlet based application in order to load resources
(3 answers)
Closed 6 years ago.
I currently have a bunch of images in my .war file like this.
WAR-ROOT
-WEB-INF
-IMAGES
-image1.jpg
-image2.jpg
-index.html
When I generate html via my servlets/jsp/etc I can simple link to
http://host/contextroot/IMAGES/image1.jpg
and
http://host/contextroot/IMAGES/image1.jpg
Not I am writing a servlet that needs to get a filesystem reference to these images (to render out a composite .pdf file in this case). Does anybody have a suggestion for how to get a filesystem reference to files placed in the war similar to how this is?
Is it perhaps a url I grab on servlet initialization? I could obviously have a properties file that explicitly points to the installed directory but I would like to avoid additional configs.
If you can guarantee that the WAR is expanded, then you can use ServletContext#getRealPath() to convert a relative web path to an absolute disk file system which you can further use in the usual Java IO stuff.
String relativeWebPath = "/IMAGES/image1.jpg";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath);
InputStream input = new FileInputStream(file);
// ...
However, if you can't guarantee that the WAR is expanded (i.e. all resources are still packaged inside WAR) and you're actually not interested on the absolute disk file system path and all you actually need is just an InputStream out of it, then use getServletContext().getResourceAsStream() instead.
String relativeWebPath = "/IMAGES/image1.jpg";
InputStream input = getServletContext().getResourceAsStream(relativeWebPath);
// ...
See also:
getResourceAsStream() vs FileInputStream
Use the getRealPath method of ServletContext.
Ex:
String path = getServletContext().getRealPath("WEB-INF/static/img/myfile.jpeg");
This is relatively straight forward you simply use the class loader to fetch the files from the class plath. :
InputStream is = YourServlet.class.getClassLoader().getResourceAsStream("IMAGES/img1.jpg");
There are a few other getResoruce classes that are worth looking at. Also you don't have to fetch the class loader through the class variable on your servlet. Any class that you happen to know has been loaded by the container should work .
If you know the relative location of the files you could ask the runtime about the exact location using
Thread.currentThread().getContextClassLoader().getResource(<relative-path>/<filename>)
This would give you an URL to the location where the specified image can be found. This URL can be used to read the specified file or you can split it to use the different parts of the URL for further processing.