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I have to read a file containing a list of strings. I'm trying to follow the advice in this post. Both solutions require using FileUtils.readLines, but use a String, not a File as the parameter.

Set<String> lines = new HashSet<String>(FileUtils.readLines("foo.txt"));

I need a File.

This post would be my question, except the OP was dissuaded from using files entirely. I need a file if I want to use the Apache method, which is the my preferred solution to my initial problem.

My file is small (a hundred lines or so) and a singleton per program instance, so I do not need to worry about having another copy of the file in memory. Therefore I could use more basic methods to read the file, but so far it looks like FileUtils.readLines could be much cleaner. How do I go from resource to file.

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    If it is a classpath resource you should not be using the File type. Consider using Google's Guava library Resources.readLines method. Commented Aug 28, 2013 at 21:33

3 Answers 3

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Apache Commons-IO has an IOUtils class as well as a FileUtils, which includes a readLines method similar to the one in FileUtils.

So you can use getResourceAsStream or getSystemResourceAsStream and pass the result of that to IOUtils.readLines to get a List<String> of the contents of your file:

List<String> myLines = IOUtils.readLines(ClassLoader.getSystemResourceAsStream("my_data_file.txt"));
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lol, so another question accepting an answer that bypasses the getting File problem. Java really doesn't like File apparently.
Has nothing to do with it. File is not system agnostic, Stream is.
@djechlin a File is an actual file in the filesystem that has a path. Resources can be part of a jar file and you can't refer to just a part of a jar (or zip) through a path. What you can do is to get a stream to either a real file or a stream to a compressed file in an archive that also unzips on the fly.
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I am assuming the file you want to read is a true resource on your classpath, and not simply some arbitrary file you could just access via new File("path_to_file");.

Try the following using ClassLoader, where resource is a String representation of the path to your resource file in your class path.

Valid String values for resource can include:

  • "foo.txt"
  • "com/company/bar.txt"
  • "com\\company\\bar.txt"
  • "\\com\\company\\bar.txt"

and path is not limited to com.company

Relevant code to get a File not in a JAR:

File file = null; try { URL url = null; ClassLoader classLoader = {YourClass}.class.getClassLoader(); if (classLoader != null) { url = classLoader.getResource(resource); } if (url == null) { url = ClassLoader.getSystemResource(resource); } if (url != null) { try { file = new File(url.toURI()); } catch (URISyntaxException e) { file = new File(url.getPath()); } } } catch (Exception ex) { /* handle it */ } // file may be null

Alternately, if your resource is in a JAR, you will have to use Class.getResourceAsStream(resource); and cycle through the file using a BufferedReader to simulate the call to readLines().

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Should resource be the string, e.g. "/com/mysite/app/file.txt"?
No, not with the preceding /. I've tested it on Windows 7 and updated with my successful test cases. Does not work with . as a path separator.
After reviewing stackoverflow.com/questions/676097/java-resource-as-file , one of the alternate answers there is similar, but not as wide in scope, as my answer.
Will new File(url.toURI()) work for resources in JAR files?
That was exactly what I was JUST going to update with; might have to use getResourceAsStream() and a BufferedReader in place of a call to FileUtils.
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using a resource to read the file to a string:

String contents = FileUtils.readFileToString( new File(this.getClass().getResource("/myfile.log").toURI()));

using inputstream:

List<String> listContents = IOUtils.readLines( this.getClass().getResourceAsStream("/myfile.log"));
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