How to read files as resource into `RandomAccessFile`?

Question

I have a jar file which contains a bunch of files in /dict/ folder. So I added it as a maven dependency. Then in order to create RandomAccessFile for a file, I read it as a resource, put into File, and then give it to the RandomAccessFile constructor. Here is how things are done:

URL resourceURL = getClass().getResource("/dict/index.verb" );
System.out.println("----> file " + resourceURL.getFile());
File f = new File(resourceURL.getFile());
System.out.println("Can read = " + f.canRead());
try {
    RandomAccessFile _file = new RandomAccessFile(f, "r");
    System.out.println(_file.length());
} catch (java.io.IOException e) {
    e.printStackTrace();
}

Here is the output:

----> modified file file:/Users/i-danielk/.m2/repository/edu/illinois/cs/cogcomp/wordnet/1.0/wordnet-1.0.jar!/dict/index.verb
Can read = false
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:497)
    at junit.framework.TestCase.runTest(TestCase.java:168)
    at junit.framework.TestCase.runBare(TestCase.java:134)
    at junit.framework.TestResult$1.protect(TestResult.java:110)
    at junit.framework.TestResult.runProtected(TestResult.java:128)
    at junit.framework.TestResult.run(TestResult.java:113)
    at junit.framework.TestCase.run(TestCase.java:124)
    at junit.framework.TestSuite.runTest(TestSuite.java:243)
    at junit.framework.TestSuite.run(TestSuite.java:238)
    at org.junit.internal.runners.JUnit38ClassRunner.run(JUnit38ClassRunner.java:83)
    at org.junit.runner.JUnitCore.run(JUnitCore.java:157)
    at com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:78)
    at com.intellij.rt.execution.junit.JUnitStarter.prepareStreamsAndStart(JUnitStarter.java:212)
    at com.intellij.rt.execution.junit.JUnitStarter.main(JUnitStarter.java:68)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:497)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:140)
java.io.FileNotFoundException: file:/Users/i-danielk/.m2/repository/edu/illinois/cs/cogcomp/wordnet/1.0/wordnet-1.0.jar!/dict/index.verb (No such file or directory)
    at java.io.RandomAccessFile.open0(Native Method)

An issue I noticed is that, using File(...) is not a good idea for reading resources.

But now I am not sure what is the best we to read the do the whole procedure without having File as an intermediate step.

Resources are not files, they are not held in file systems, and you cannot perform random access on them. You can get them as URLs or streams, period. — user207421, Nov 16 '15 at 02:58

score 2 · Answer 1 · answered Nov 16 '15 at 02:48

2

URL#getFile isn't doing what you think it is and you should read the JavaDocs to find out what it does do.

Instead, you should use something like URL#openStream and write the contents out to a physical File yourself.

As a rough example...

URL resourceURL = getClass().getResource("/dict/index.verb");
File output = new File("some file somewhere");
try (InputStream is = resourceURL.openStream(); OutputStream os = new FileOutputStream(output)) {
    byte[] buffer = new byte[2048];
    int bytesRead = -1;
    while ((bytesRead = is.read(buffer)) != -1) {
        os.write(buffer, 0, bytesRead);
    }
} catch (IOException exp) {
    exp.printStackTrace();
}

You might find File.createTempFile of some use

answered Nov 16 '15 at 02:48

MadProgrammer

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So I should read the resource, write it, and then read it into `RandomAccessFile`? Isn't it inefficient? – Daniel Nov 16 '15 at 03:01
Your resource is a series of compressed bytes in side a Jar (zip) file (with a bunch of other compressed bytes). You can read it like you would a normal file, because it's not. Instead, you need to extract the contents to a file system file and then read it. Also remember, you won't be able to write to this resource (you should accept that for now because it's a complete pain to update a running Jar file which fails more often then it succeeds) – MadProgrammer Nov 16 '15 at 03:07

How to read files as resource into `RandomAccessFile`?

1 Answers1