如何從Java中的數據庫檢索JSON對象

Question

我嘗試使用傑克遜獲取json對象，但得到如下異常：

SEVERE: Servlet.service() for servlet [JsonProcessor] in context with path [/JJS2] threw exception [Servlet execution threw an exception] with root cause
java.lang.NoSuchMethodError: com.fasterxml.jackson.core.JsonGenerator.setCurrentValue(Ljava/lang/Object;)V

這是Java代碼：

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

        JSONObject jsonResponse = new JSONObject();
         List<Map<String, Object>> result = new ArrayList<>();// JDK7++

        try {
            JSONArray data = new JSONArray();
            Connection con = OracleDBConnection.getConnection();
            String query = "Select * from JJS";
            Statement st = con.createStatement();
            ResultSet rSet = st.executeQuery(query);

            while (rSet.next()) {
                Map<String, Object> row = new HashMap<>();
                row.put("JSON_Diagram", rSet.getString("JSON_INFO"));
                result.add(row);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

        ObjectMapper mapper = new ObjectMapper();
        try {
            response.getWriter().write(mapper.writeValueAsString(result));
            response.getWriter().flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

這是數據庫中的json示例：

{"cells":
[{"type":"basic.Rect","position":{"x":-2,"y":33},
 "size":{"width":71,"height":625},"angle":0,"isInteractive":false,
 "id":"a55845b6-b753-42f0-b361-f0fcd3eaa611","z":1,
 "attrs":{"rect":{"fill":"#EEEEEE","stroke":"#008B8B","stroke-width":2},
".":{"magnet":false}}}

該數據庫只有一列包含json數據。

Answer 1

您在jackson-core （流，低級編碼器/解碼器）和jackson-databind （對象映射）之間存在版本不兼容。 您很可能擁有2.6的jackson-databind ，但有一些早期版本的jackson-core 。 這些組件的次要版本必須匹配； 或至少jackson-databind次要版本不能晚於jackson-core版本。