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java - 在 Java 中，在 Android 上将 2D pojo 数组序列化为 json

问题描述

我正在构建一个国际象棋游戏，我想序列化棋盘并在每次移动后将其发送到服务器。我将使用 JSON 来完成这项任务。

我的游戏的表示方式是我创建的自定义 POJO 的二维数组，名为Piece.

目前我只是在创建一个 JSON 风格的字符串，而不是实际的 JSON，这是有问题的，因为我现在不能像对待真正的 JSON 一样对待它，对它进行索引等等。

这是我用来创建这个 JSON 结构的代码：

StringBuilder result = new StringBuilder();
for (int i = 0; i < piecesMatrix.length; i++) {
    result.append("\"row " + i + "\":\n");
    for (int j = 0; j < piecesMatrix[i].length; j ++) {
        result.append("\"column " + j + "\": ");
        if (piecesMatrix[i][j] == null || piecesMatrix[i][j] instanceof LegalMove) {
            result.append(" \"null\",\n");
        } else {
            result.append(" \"" + piecesMatrix[i][j].getClass().getSimpleName() + "\",\n");
        }
    }
}
message = result.toString();

结果呈现如下：

[ 10-06 20:35:19.971  7509: 7509 D/Data = 

 ]
"row 0":
"column 0":  "BlackRook",
"column 1":  "BlackKnight",
"column 2":  "BlackBishop",
"column 3":  "BlackQueen",
"column 4":  "BlackKing",
"column 5":  "BlackBishop",
"column 6":  "BlackKnight",
"column 7":  "BlackRook",
"row 1":
"column 0":  "BlackPawn",
"column 1":  "BlackPawn",
"column 2":  "BlackPawn",
"column 3":  "BlackPawn",
"column 4":  "BlackPawn",
"column 5":  "BlackPawn",
"column 6":  "BlackPawn",
"column 7":  "BlackPawn",
"row 2":
"column 0":  "null",
"column 1":  "null",
"column 2":  "null",
"column 3":  "null",
"column 4":  "null",
"column 5":  "null",
"column 6":  "null",
"column 7":  "null",
"row 3":
"column 0":  "null",
"column 1":  "null",
"column 2":  "null",
"column 3":  "null",
"column 4":  "null",
"column 5":  "null",
"column 6":  "null",
"column 7":  "null",
"row 4":
"column 0":  "null",
"column 1":  "WhitePawn",
"column 2":  "null",
"column 3":  "null",
"column 4":  "null",
"column 5":  "null",
"column 6":  "null",
"column 7":  "null",
"row 5":
"column 0":  "null",
"column 1":  "null",
"column 2":  "null",
"column 3":  "null",
"column 4":  "null",
"column 5":  "null",
"column 6":  "null",
"column 7":  "null",
"row 6":
"column 0":  "WhitePawn",
"column 1":  "null",
"column 2":  "WhitePawn",
"column 3":  "WhitePawn",
"column 4":  "WhitePawn",
"column 5":  "WhitePawn",
"column 6":  "WhitePawn",
"column 7":  "WhitePawn",
"row 7":
"column 0":  "WhiteRook",
"column 1":  "WhiteKnight",
"column 2":  "WhiteBishop",
"column 3":  "WhiteQueen",
"column 4":  "WhiteKing",
"column 5":  "WhiteBishop",
"column 6":  "WhiteKnight",
"column 7":  "WhiteRook",

问题是，我怎样才能把它变成真正的 JSON？

理想情况下，我希望输出如下所示：

"row 0":
    "column 0": "Rook",
    "column 1": "Knight",
    "column 2": "Bishop",
    ...,
"row 1":
    "column 0": "Pawn",
    "column 1": "null",
    "column 2": "Pawn",
    ...,
"row 2":
    "column 0": "null",
    "column 1": "Pawn",
    "column 2": "null",
    ...

编辑：

我试过了：

Gson gson = new 
GsonBuilder().setPrettyPrinting().serializeNulls().create();
    message = gson.toJson(piecesMatrix);

我有：

SecurityException: Can not make a java.lang.reflect.Method constructor accessible

编辑二

我试过了：

Gson gson = new GsonBuilder().setPrettyPrinting().serializeNulls().create();
    message = gson.toJson(piecesMatrix, Piece.class);

我有：

    [ 10-07 00:28:20.591 10899:10899 D/Data = 

 ]
{}

标签： javaandroidjsongson

解决方案

是这样的吗……

        JSONArray arr = new JSONArray();
    for (int i = 0; i < piecesMatrix.length; i++) {
        JSONArray innerArray = new JSONArray();

        for (int j = 0; j < piecesMatrix[i].length; j++) {
            JSONObject column = new JSONObject();

            if (piecesMatrix[i][j] == null || piecesMatrix[i][j] instanceof LegalMove) {
                try {
                    column.put("column " + String.valueOf(j), null);
                } catch (Exception e) {}
            } else {
                try {
                    column.put("column " + String.valueOf(j), piecesMatrix[i][j].getClass().getSimpleName());
                } catch (Exception e) {}
            }
            innerArray.put(column);
        }
        JSONObject json = new JSONObject();
        try {
            json.put("row " + i, innerArray);
        } catch (JSONException e) {}
        arr.put(json);
    }

    Log.d("Data = \n\n", arr.toString());

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