Question

对于一个项目，我们正在使用一个telnet服务器来接收我们的数据。从telnet连接读取一行后，我得到以下字符串：

原始字符串：

SVR GAME MOVE {PLAYER: "PLAYER", MOVE: "1", DETAILS: ""}

我试图将其转换为输出，我可以轻松使用而不会发生太多可利用性。所需的格式是轻松获得（PLAYER）或获得。我尝试将regex与json结合使用，请参阅下面的代码：

String line = currentLine;
line = line.replaceAll("(SVR GAME MOVE )", ""); //remove SVR GAME MATCH
line = line.replaceAll("(\"|-|\\s)", "");//remove quotations and - and spaces (=\s)
line = line.replaceAll("(\\w+)", "\"$1\""); //add quotations to every word
JSONParser parser = new JSONParser();
try {
       JSONObject json = (JSONObject) parser.parse(line);
       //@todo bug when details is empty causes
       //@todo example string: SVR GAME MOVE {PLAYER: "b", MOVE: "1", DETAILS: ""}
       //@todo string that (line) that causes an error when parsing to json {"PLAYER":"b","MOVE":"1","DETAILS":}
       //@todo Unexpected token RIGHT BRACE(}), i think because "details" has no value
       System.out.println(json.get("PLAYER"));
       System.out.println(json.get("MOVE"));
       System.out.println(json.get("DETAILS"));
       int index = Integer.valueOf(json.get("MOVE").toString());
       for(GameView v : views){
          v.serverMove(index);//dummy data is index 1
       }
    } catch (ParseException e) {
        e.printStackTrace();
}

问题是当细节为空时，这将导致意外的令牌RIGHT BRACE（}）。此外，当玩家在其名称中使用剥削性名称（例如引号）时，代码将很容易崩溃。

将原始字符串转换为输出的最佳方法是什么？您可以轻松获得单独的设置（播放器，移动，详细信息）？

Answer 1

这将解析字符串并将变量放在地图中：

    Map<String,String> vars = new HashMap<>();
    String input = "SVR GAME MOVE {PLAYER: \"PLAYER\", MOVE: \"1\", DETAILS: \"\"}";
    Pattern pattern = Pattern.compile("([A-Za-z]+): \"([^\"]*)\"");
    Matcher matcher = pattern.matcher(input);
    while (matcher.find()) {
        vars.put(matcher.group(1), matcher.group(2));
    }

Answer 2

使用jackson library，您可以执行以下操作：

String input = "{PLAYER: \"PLAYER\", MOVE: \"1\", DETAILS: \"\"}";

ObjectMapper mapper = new ObjectMapper();
JsonNode yourJson = mapper.readTree(input);
System.out.println(yourJson.get("PLAYER"));
System.out.println(yourJson.get("MOVE"));
System.out.println(yourJson.get("DETAILS"));

将String（使用Regex）转换为可用对象

2 个答案: