Question

我是JSON和JSP的新手。我正在编写一个REST url，它将从java swing应用程序中调用。 url将显示一些JSON数据。下面是我的jsp代码：

JSONObject MainObject=null;
  JSONObject json1=null;
  JSONArray jarr=new JSONArray();

  String email = request.getParameter("EMAIL");

  cc = new MySQLConnection();

String query = "SELECT * FROM CLIENTS WHERE EMAIL = '"+email+"'";
try{
conn =cc.db();
stmtt =conn.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE, ResultSet.CONCUR_READ_ONLY);
rs = stmtt.executeQuery(query);  
if(rs.next()){
    json1 = new JSONObject();
    MainObject = new JSONObject();

    json1.put("EMAIL", rs.getString("EMAIL"));
    json1.put("ID", rs.getLong("ID"));
    json1.put("COMPANY", rs.getString("COMPANY"));
    json1.put("CODE", rs.getString("CODE"));
    json1.put("VALID_TILL", rs.getLong("VALID_TILL"));
    jarr.put(json1);

    MainObject.put("SUCCESS", "1");
    MainObject.put("CLIENTS", jarr); 

    }
//else{
//out.println("Error....");
//}
    out.println(MainObject);
}

catch(SQLException ex){out.println(ex);}
finally{try{stmtt.close(); rs.close();conn.close();} catch(SQLException ex){out.println(ex);} }

这给了我以下输出：{"SUCCESS":"1","CLIENTS":[{"VALID_TILL":20,"COMPANY":"BOOK PALACE","ID":1,"EMAIL":"pranjal","CODE":"98877655"}]}

现在，当我尝试解析Swing应用程序中的数据时：它给了我错误。我正在解析这个：

 List<NameValuePair> params = new ArrayList<>();
           params.add(new BasicNameValuePair("EMAIL", mail)) ;
            JSONObject json = jParser.makeHttpRequest(YOURUrl, "GET", params);
            System.out.println(json.toString());
             try {
                int success = json.getInt("SUCCESS");                

                if (success == 1) {
                    products = json.getJSONArray("CLIENTS");                   
                    for (int i = 0; i < products.length(); i++) {
                        JSONObject c = products.getJSONObject(i);

                        String Company = c.getString("COMPANY");
                        String Name = c.getString("C_NAME");
                        String Email = c.getString("EMAIL");
                        String Code = c.getString("CODE");
                        String ValidTill = c.getString("VALID_TILL");
                        String sl = c.getString("ID");
                        COMPANY = Company;
                        CNAME = Name;
                        EMAIL = Email;
                        CODE = Code;
                        VALID_TILL = ValidTill;
                        SL = sl;

                    }
                } 
            } catch (JSONException e) {
                System.err.println(e);
            }

我的jsp文件中有什么错误？我在哪里做错了？

JSONObject文本必须以＆＃39; {＆＃39; 1点

0 个答案: