How to chain multiple different InputStreams into one InputStream

41,005

Solution 1

It's right there in JDK! Quoting JavaDoc of SequenceInputStream:

A SequenceInputStream represents the logical concatenation of other input streams. It starts out with an ordered collection of input streams and reads from the first one until end of file is reached, whereupon it reads from the second one, and so on, until end of file is reached on the last of the contained input streams.

You want to concatenate arbitrary number of InputStreams while SequenceInputStream accepts only two. But since SequenceInputStream is also an InputStream you can apply it recursively (nest them):

new SequenceInputStream(
    new SequenceInputStream(
        new SequenceInputStream(file1, file2),
        file3
    ),
    file4
);

...you get the idea.

See also

Solution 2

This is done using SequencedInputStream, which is straightforward in Java, as Tomasz Nurkiewicz's answer shows. I had to do this repeatedly in a project recently, so I added some Scala-y goodness via the "pimp my library" pattern.

object StreamUtils {
  implicit def toRichInputStream(str: InputStream) = new RichInputStream(str)

  class RichInputStream(str: InputStream) {
// a bunch of other handy Stream functionality, deleted

    def ++(str2: InputStream): InputStream = new SequenceInputStream(str, str2)
  }
}

With that, I can do stream sequencing as follows

val mergedStream = stream1++stream2++stream3

or even

val streamList = //some arbitrary-length list of streams, non-empty
val mergedStream = streamList.reduceLeft(_++_)

Solution 3

Another solution: first create a list of input stream and then create the sequence of input streams:

List<InputStream> iss = Files.list(Paths.get("/your/path"))
        .filter(Files::isRegularFile)
        .map(f -> {
            try {
                return new FileInputStream(f.toString());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }).collect(Collectors.toList());

new SequenceInputStream(Collections.enumeration(iss)))

Solution 4

Here is a more elegant solution using Vector, this is for Android specifically but use vector for any Java

    AssetManager am = getAssets();
    Vector v = new Vector(Constant.PAGES);
    for (int i =  0; i < Constant.PAGES; i++) {
        String fileName = "file" + i + ".txt";
         InputStream is = am.open(fileName);
         v.add(is);
    }
    Enumeration e = v.elements();
    SequenceInputStream sis = new SequenceInputStream(e);
    InputStreamReader isr = new InputStreamReader(sis);

    Scanner scanner = new Scanner(isr);   // or use bufferedReader
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41,005
Magnus
Author by

Magnus

Updated on June 13, 2020

Comments

  • Magnus
    Magnus almost 4 years

    I'm wondering if there is any ideomatic way to chain multiple InputStreams into one continual InputStream in Java (or Scala).

    What I need it for is to parse flat files that I load over the network from an FTP-Server. What I want to do is to take file[1..N], open up streams and then combine them into one stream. So when file1 comes to an end, I want to start reading from file2 and so on, until I reach the end of fileN.

    I need to read these files in a specific order, data comes from a legacy system that produces files in barches so data in one depends on data in another file, but I would like to handle them as one continual stream to simplify my domain logic interface.

    I searched around and found PipedInputStream, but I'm not positive that is what I need. An example would be helpful.

  • Magnus
    Magnus over 11 years
    Doh! I must have really bad luck with my keywords when googling today. Because I gave it a good googling. I figured that I can't really be the first one needing this, thanks!
  • adi
    adi over 10 years
    SequenceInputStream constructor also accepts a Enumeration, so you don't have to do the nesting.
  • pretzels1337
    pretzels1337 over 10 years
    Are they any concerns if I chain hundreds of inputstreams using this technique?
  • Thierry
    Thierry over 7 years
    @pretzels1337, a bit late. yes there is a concern : you will open all files at once, when creating the enumeration of inputstream, and there is a limit to the number of files that can be opened concurrently (set by the OS). For this use case, I would rather copy the SequenceInputStream code, replace the Enumeration<InputStream> with List<File>, and alter the nextStream() and close() methods accordingly. You could provide your own Enumeration<InputStream> that opens the stream in nextElement(), but look at what will happen in close() in case of IOExcpetion on the first inputstream
  • ed22
    ed22 about 6 years
    Can I seek to random position in SequenceInputStream? I mean forward and backward.