Question

我正在尝试使用Android中的Http请求解析我从以下URL中检索到的XML文件：

http://www.musicbrainz.org/ws/2/recording/?query=kick%20AND%20artist:inxs

解析它的最佳方法是什么？我需要找到艺术家姓名，专辑名称和发行ID。我一直在尝试使用此处指定的方法：

http://developer.android.com/reference/org/xmlpull/v1/XmlPullParser.html

但事实证明它真的很无效。有关库的建议或我可以用来检索所需信息的简单方法吗？

到目前为止我的代码如下：

public List parse(InputStream in) throws XmlPullParserException, IOException {
    try {
        XmlPullParser parser = Xml.newPullParser();
        parser.setFeature(XmlPullParser.FEATURE_PROCESS_NAMESPACES, false);
        parser.setInput(in, null);
        parser.nextTag();
        return readFeed(parser);
    } finally {
        in.close();
    }
}

private List readFeed(XmlPullParser parser) throws XmlPullParserException, IOException {
    List entries = new ArrayList();

    parser.require(XmlPullParser.START_TAG, ns, "metadata");
    Log.i("XMLParser", "metadata");
    while (parser.next() != XmlPullParser.END_TAG) {
        if (parser.getEventType() != XmlPullParser.START_TAG) {
            continue;
        }
        String name = parser.getName();
        // Starts by looking for the entry tag
        if (name.equals("recording-list")) {
            Log.i("XMLParser", "recording-list");
            entries.add(readRecordingList(parser));
        } else {
            skip(parser);
        }
    }  
    return entries;
}

private Recording readRecordingList(XmlPullParser parser) throws XmlPullParserException, IOException {
    parser.require(XmlPullParser.START_TAG, ns, "recording-list");
    Recording recording=new Recording(null,null,null,null);

    while (parser.next() != XmlPullParser.END_TAG) {
        if (parser.getEventType() != XmlPullParser.START_TAG) {
            continue;
        }
        String name = parser.getName();
        if (name.equals("recording")) {
            Log.i("XMLParser", "recording");
            recording = readRecording(parser);
        } else {
            skip(parser);
        }
    }
    return recording;
}

private Recording readRecording(XmlPullParser parser) throws XmlPullParserException, IOException {
    parser.require(XmlPullParser.START_TAG, ns, "recording");
    Recording recording=new Recording(null,null,null,null);
    String albumTitle=null;
    String artistName=null;

    while (parser.next() != XmlPullParser.END_TAG) {
        if (parser.getEventType() != XmlPullParser.START_TAG) {
            continue;
        }
        String name = parser.getName();
        Log.i("XMLParserNodeName", name);
        if (name.equals("title")) {
            Log.i("XMLParser", "title");
            albumTitle=readTitle(parser);
            Log.i("XMLParser", albumTitle);
            //recording = readRecording(parser);
        }else if(name.equals("name")){ 
            Log.i("XMLParser", "name");
            name=readArtistName(parser);
            Log.i("XMLParser", name);
        }else {
            skip(parser);
        }
    }
    return recording;
}


private void skip(XmlPullParser parser) throws XmlPullParserException, IOException {
    if (parser.getEventType() != XmlPullParser.START_TAG) {
        throw new IllegalStateException();
    }
    int depth = 1;
    while (depth != 0) {
        switch (parser.next()) {
        case XmlPullParser.END_TAG:
            depth--;
            break;
        case XmlPullParser.START_TAG:
            depth++;
            break;
        }
    }
 }

private String readText(XmlPullParser parser) throws IOException, XmlPullParserException {
    String result = "";
    if (parser.next() == XmlPullParser.TEXT) {
        result = parser.getText();
        parser.nextTag();
    }
    return result;
}

private String readTitle(XmlPullParser parser) throws IOException, XmlPullParserException {
    parser.require(XmlPullParser.START_TAG, ns, "title");
    String title = readText(parser);
    parser.require(XmlPullParser.END_TAG, ns, "title");
    return title;
}

private String readArtistName(XmlPullParser parser) throws IOException, XmlPullParserException {
    parser.require(XmlPullParser.START_TAG, ns, "name");
    String name = readText(parser);
    parser.require(XmlPullParser.END_TAG, ns, "name");
    return name;
}

public static class Recording {


        public final String recordingTitle;
        public final String releaseTitle;
        public final String releaseID;
        public final String artistName;

        public Recording(String recordingTitle, String releaseTitle,
                String releaseID, String artistName) {
            this.recordingTitle = recordingTitle;
            this.releaseTitle = releaseTitle;
            this.releaseID = releaseID;
            this.artistName = artistName;
        }    
}

}

基本上这是Android中XML Pull Parser Tutorial的改编版，但我想避免在每个节点编写这么多代码，而xml文件中有很多代码。我认为我更喜欢XML Pull Parser而不是DOM，因此我不会消耗大量内存。

Answer 1

Documentation of the MusicBrainz Web Service提到了一个可以使用的java库：musicbrainzws2-java。

我没有使用该特定库的经验，但这些库倾向于提供API以透明地请求来自MusicBrainz的实体。选择正确的终点和解析是由库完成的。

您不是第一个尝试使用带有Java的MusicBrainz Web服务的人; - ）

wiki of the library有一些关于如何使用它进行搜索的示例。

解析此XML文件的最佳方法

1 个答案: