为什么我的匹配器失败了？

Question

我正在将一个字符串传递给我的歌曲解析器方法,它失败了,我无法弄清楚为什么.每件事都返回null或0.

我的解析器方法是

 public static Song parseSong(String songString){
  Map<String, String> songMap = new HashMap<String, String>();
  Pattern pattern = Pattern.compile(".*<key>(.+)</key><(.+)>(.+)</.+>.*\n");
  Scanner scanner = new Scanner(songString);
  if(scanner.hasNext(pattern))
  {
     String line = scanner.next(pattern);
     Matcher matcher = pattern.matcher(line);
     MatchResult result = matcher.toMatchResult();
     songMap.put(result.group(1), result.group(3));
  }
  int count = 0, rating = 0;
  try{
     count = Integer.parseInt(songMap.get("Play Count"));
  }
  catch(Exception e)
  {
     //bury this for now will handle when rest is working
  }
  try{
     rating = Integer.parseInt(songMap.get("Rating"));
  }
  catch(Exception e)
  {
     //bury this for now will handle when rest is working
  }
  return new Song(songMap.get("Name"), songMap.get("Artist"), songMap.get("Album"),
        songMap.get("Genre"), count, rating, songMap.get("Location"));

}

      String songString = "<key>Track ID</key><integer>160</integer>\n"+
     "<key>Name</key><string>Ashley</string>\n"+
    " <key>Artist</key><string>Escape the Fate</string>\n"+
    " <key>Composer</key><string>Luca Gusella</string>\n"+
    " <key>Album</key><string>This War Is Ours</string>\n"+
  "   <key>Genre</key><string>Metal</string>\n"+
     "<key>Kind</key><string>AAC audio file</string>\n"+
  "  <key>Size</key><integer>7968219</integer>\n"+
   "  <key>Total Time</key><integer>246503</integer>\n"+
  "   <key>Track Number</key><integer>17</integer>\n"+
   "  <key>Year</key><integer>2005</integer>\n"+
   "  <key>Date Modified</key><date>2009-07-27T01:17:29Z</date>\n"+
    " <key>Date Added</key><date>2009-07-27T01:17:00Z</date>\n"+
    "<key>Play Count</key><integer>150</integer>\n"+
    " <key>Bit Rate</key><integer>256</integer>\n"+
    " <key>Sample Rate</key><integer>44100</integer>\n"+
    " <key>Comments</key><string>\"Amanda\" performed by Aisha Duo from the CD Quiet Songs, courtesy of Obliq Sound.  Written by Luca Gusella, published by Editions ObliqMusic (GEMA).  All Rights Reserved.  Used by Permission. </string>\n"+
    " <key>Skip Count</key><integer>1</integer>\n"+
    " <key>Skip Date</key><date>2009-07-27T01:46:32Z</date>\n"+
    " <key>Artwork Count</key><integer>1</integer>\n"+
    " <key>Persistent ID</key><string>A4D6F35FE9F41B58</string>\n"+
    " <key>Track Type</key><string>File</string>\n"+
    " <key>Location</key><string>file://localhost/C:/Documents%20and%20Settings/MB24244/Desktop/music/07%20Knees.m4a</string>\n"+
     "<key>File Folder Count</key><integer>4</integer>\n"+
     "afgjdhfshsgsughghanoise\n"+
     "<key>Library Folder Count</key><integer>1</integer>\n"+
     "<key>Rating</key><integer>100</integer>";

任何人都可以帮助解释我的方法有什么问题以及为什么这些小组不工作(这似乎是问题所在)

Answer 1

为什么不使用XML解析器来解析XML？

虽然查看XML示例但它不是很好,因为它实际上是建模map而不是建模<song>

看看你的正则表达式为什么你要找行结尾\n.您似乎依次匹配每一行,我不相信这些将包含新行字符.

但是,这种不使用扫描仪的方法有效.请注意,我已更改正则表达式以删除行结尾.

    Map<String, String> songMap = new HashMap<String, String>();

    Pattern pattern = Pattern
            .compile(".*<key>(.+)</key><(.+)>(.+)</.+>.*");

    String[] lines = songString.split("\n");

    for (String line : lines) {
        Matcher matcher = pattern.matcher(line);
        if (matcher.matches()) {
            songMap.put(matcher.group(1), matcher.group(3));
        }
    }

您也可以使用扫描仪.