如何从mysql返回JSON对象？

我正在尝试设置此对象：

var userPref = {} 

从此查询返回到mysql：

var qSelect = "SELECT age_min, age_max, max_distance, partner_gender_id, partner_children, partner_family_plans FROM User_Preferences WHERE user_id = ?";
            var qValues = [user_pref_id];
            var qCall = mysql.format(qSelect, qValues);
            connection.query(qCall, function(err, userPrefFound, fields) {
                if (err) {
                    console.log("get my preferences error: " + err);
                    callback(false);
                } else {
                    userPref = userPrefFound[0];
                    console.log("userPref", userPref)
                    if (userPrefFound.length > 0 && userPref.max_distance < 300) {
                        getUserLocation();
                    } else {
                        getUserPrefEth();

                    }
                }
            });

我的userPref打印为RowDataPacket

userPref RowDataPacket {
  age_min: 18,
  age_max: 65,
  max_distance: 5,
  partner_gender_id: 2,
  partner_children: 2,
  partner_family_plans: 3 }

然后不允许我像这样设置对象

var politics_interested = [];
        var qSelect = "SELECT politics_type FROM User_Pref_Politics WHERE user_id = ?";
        var qValues = [user_pref_id];
        var qCall = mysql.format(qSelect, qValues);
        connection.query(qCall, function(err, userPrefPolitics, fields) {
            if (err) {
                console.log("get my preferences politics error: " + err);
                callback(false);
            } else {
                if (userPrefPolitics.length > 0) {
                    userPrefPolitics.forEach(function(i) {

                        politics_interested.push(i.politics_type);
                    });
                }
                console.log("politics_interested", userPref)
                userPref["politics_interested"] = politics_interested;
                getUserPrefMarital();

            }
        });

并显示此错误：

\node_modules\mysql\lib\protocol\Parser.js:437
      throw err; // Rethrow non-MySQL errors
      ^

TypeError: Cannot set property 'politics_interested' of undefined

我该如何从mysql获取返回值并将其设置为对象？

我尝试了一个数组userPref ['religious_level_interested'] = JSON.parse（JSON.stringify（userPrefRelig））;但是，当我尝试将其用于某个对象时，它将无法正常工作，不确定是否可以正常工作。