Question

我想用jFugue在applet中播放一些MIDI音乐。 MIDI模式有一个类 - Pattern - 加载模式的唯一方法是从文件中获取。现在，我不知道applet如何加载文件和什么不是，但我使用的框架（PulpCore）使加载资产成为一项简单的任务。如果我需要从ZIP目录中获取资产，我可以使用提供Assets和get()方法的getAsStream()类。 get()将给定资源作为ByteArray返回，另一个作为InputStream返回。

我需要jFugue从ByteArray或InputStream加载模式。在伪代码中，我想这样做：

Pattern.load(new File(Assets.get("mymidifile.midi")));

但是没有File构造函数可以使用ByteArray。建议，拜托？

Answer 1

确实jFugue不允许加载除文件之外的任何内容，这是一种耻辱，因为没有什么可以阻止使用任何其他类型的流：

public static final String TITLE = "Title";

public static Pattern loadPattern(File file) throws IOException {
    InputStream in = new FileInputStream(file);
    try {
        return loadPattern(in);
    } finally {
        in.close();
    }
}

public static Pattern loadPattern(URL url) throws IOException {
    InputStream in = url.openStream();
    try {
        return loadPattern(in);
    } finally {
        in.close();
    }
}

public static Pattern loadPattern(InputStream in) throws IOException {
    return loadPattern(new InputStreamReader(in, "UTF-8")); // or ISO-8859-1 ?
}

public static Pattern loadPattern(Reader reader) throws IOException {
    if (reader instanceof BufferedReader) {
        return loadPattern(reader);
    } else {
        return loadPattern(new BufferedReader(reader));
    }
}

public static Pattern loadPattern(BufferedReader bread) throws IOException {
    StringBuffer buffy = new StringBuffer();

    Pattern pattern = new Pattern();
    while (bread.ready()) {
        String s = bread.readLine();
        if ((s != null) && (s.length() > 1)) {
            if (s.charAt(0) != '#') {
                buffy.append(" ");
                buffy.append(s);
            } else {
                String key = s.substring(1, s.indexOf(':')).trim();
                String value = s.substring(s.indexOf(':')+1, s.length()).trim();
                if (key.equalsIgnoreCase(TITLE)) {
                    pattern.setTitle(value);
                } else {
                    pattern.setProperty(key, value);
                }
            }
        }
    }
    return pattern;
}

UPDATE（对于loadMidi）

public static Pattern loadMidi(InputStream in) throws IOException, InvalidMidiDataException
{
    MidiParser parser = new MidiParser();
    MusicStringRenderer renderer = new MusicStringRenderer();
    parser.addParserListener(renderer);
    parser.parse(MidiSystem.getSequence(in));
    Pattern pattern = new Pattern(renderer.getPattern().getMusicString());
    return pattern;
}

public static Pattern loadMidi(URL url) throws IOException, InvalidMidiDataException
{
    MidiParser parser = new MidiParser();
    MusicStringRenderer renderer = new MusicStringRenderer();
    parser.addParserListener(renderer);
    parser.parse(MidiSystem.getSequence(url));
    Pattern pattern = new Pattern(renderer.getPattern().getMusicString());
    return pattern;
}

Answer 2

如果我没错，Pattern文件包含纯文本。使用getAsStream（）加载文件，然后使用

将其转换为字符串

BufferedReader br = new BufferedReader(new InputStreamReader(yourStream));
//...
String pattern = convertToString(br); // you should implement convertToString yourself. It's easy. Read the java.io APIs.

其中yourStream是getAsStream（）返回的InputStream。然后使用add（String ... patterns）方法加载模式：

add(pattern);

Answer 3

您可以使用此代码（取自Pattern.loadPattern()方法的实现）：

    InputStream is = ...; // Get a stream from the Asset object

    // Prepare a pattern object
    Pattern pattern = new Pattern();

    // Now start reaing from the stream
    StringBuffer buffy = new StringBuffer();
    BufferedReader bread = new BufferedReader(new InputStreamReader(is));
    while (bread.ready()) {
        String s = bread.readLine();
        if ((s != null) && (s.length() > 1)) {
            if (s.charAt(0) != '#') {
                buffy.append(" ");
                buffy.append(s);
            } else {
                String key = s.substring(1, s.indexOf(':')).trim();
                String value = s.substring(s.indexOf(':')+1, s.length()).trim();
                if (key.equalsIgnoreCase(TITLE)) {
                    pattern.setTitle(value);
                } else {
                    pattern.setProperty(key, value);
                }
            }
        }
    }
    bread.close();
    pattern.setMusicString(buffy.toString());

    // Your pattern is now ready

Answer 4

您可以读取字节数组并将其转换为字符串。

问题将是InputStream。有一个StringBufferInputStream，但它不赞成使用StringReader。

byte [] b = Assets.get();
InputStream is = new StringBufferInputStream(new String(b));
Pattern.load(is);

Answer 5

您不想使用File，而是想使用java.io.ByteArrayInputStream

Java File和ByteArray或InputStream

5 个答案: