table 单元格为空时的 Gson 空指针异常

Question

数据库table结构：

Card Number     Name   Bal  Withdrawal  Deposit  Bal        Tx Date     Usage_Type

AL98****66325   MOSES   0    12981.02    1   -362764.96    16-Oct-14       N/A
AL98****66325   MOSES   0    50000.01        -362764.96    17-Oct-14       Manual

如您所见，"Deposit" 的第二行中的值为空。它不是零或空。所以我得到异常说：

java.lang.NullPointerException atcom.google.gson.JsonPrimitive.isPrimitiveOrString(JsonPrimitive.java:278) at com.google.gson.JsonPrimitive.setValue(JsonPrimitive.java:100) at com.google.gson.JsonPrimitive.<init>(JsonPrimitive.java:65) at com.In10s.getTable.GetTable.doGet(GetTable.java:49)

我从我的客户那里得到 xml 数据，可能有很多空值，我无法通过数百个文件来替换空值。有可能解决这个问题吗？

有没有gson的替代品。我还没有尝试过杰克逊图书馆。我不知道如何将结果集转换为 jackson 中的 json 对象如果我必须转移。

Java代码：

public class GetTable extends HttpServlet {

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

        try {
            JsonObject jsonResponse = new JsonObject();
            JsonArray data = new JsonArray();
            PrintWriter out = response.getWriter();

            Connection con = OracleDBConnection.getConnection();
            String query = "Select * from CREDIT_CARD_TRANSACTIONS";
            Statement st = con.createStatement();
            ResultSet rSet = st.executeQuery(query);

            while (rSet.next()) 
            {
                JsonArray row = new JsonArray();
                row.add(new JsonPrimitive(rSet.getString("CARD_NUMBER")));
                row.add(new JsonPrimitive(rSet.getString("FIRST_NAME")));
                row.add(new JsonPrimitive(rSet.getString("OPENING_BALANCE")));
                row.add(new JsonPrimitive(rSet.getString("WITHDRAWAL")));
                row.add(new JsonPrimitive(rSet.getString("DEPOSIT")));
                row.add(new JsonPrimitive(rSet.getString("CLOSING_BAL")));
                row.add(new JsonPrimitive(rSet.getString("TXDATE")));
                row.add(new JsonPrimitive(rSet.getString("USAGE_TYPE")));
                data.add(row);
            }

            jsonResponse.add("ResponseData", data);
            response.setContentType("application/json");
            response.setCharacterEncoding("UTF-8");

            out.print(jsonResponse);
            out.flush();
            System.out.println(jsonResponse);

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

Answer 1

我建议使用 ObjectMapper. You can download it from here。

首先将您的所有响应设置为映射列表或 POJO 列表：

 List<Map<String, Object>> response = new ArrayList<>();// JDK7++
 while (rSet.next()) 
            {
                Map<String, Object> row = new HashMap<>;
                row.put("CARD_NUMBER",  rSet.getString("CARD_NUMBER"));
                // this wil be null safe
                ...
                response.add(row);
            }

当您填写回复列表时，将此列表转换为 Json 字符串：

ObjectMapper mapper = new ObjectMapper();

try {
    response.getWriter().write(mapper.writeValueAsString(result));
    out.flush();
} catch (JsonGenerationException e) {

    e.printStackTrace();

} catch (JsonMappingException e) {

    e.printStackTrace();

} catch (IOException e) {

    e.printStackTrace();

}

table 单元格为空时的 Gson 空指针异常

Gson null pointer exception when table cell is empty

java

jackson

gson