javascript - 有没有办法在 Op.op sequelize 中循环遍历许多参数的对象

Question

我正在设置一个搜索API，就是搜索； id、type、originCity、destinationCity、departmentDate、reason、accommodation、approvalStatus...，之后列表可能会增加

const options = {
        id,
        type,
        originCity,
        destinationCity,
        departureDate,
        reason,
        accommodation,
        boolApprovalStatus
      };

我正在使用 sequelize Op 运算符，我希望我的代码可重用，我不喜欢将所有这些参数转储到 sequelize Op.op 运算符中。
{ where:{ [Op.or]: [{id}, {type}, {originCity}, {destinationCity}, {departureDate}, {reason}, {accommodation}, {approvalStatus}] } }
我尝试使用没有返回任何结果的 Object.value(options) 方法，并保持我的服务器正常运行，直到超时

这是我下面的代码

import { Op } from 'sequelize';
import Request from '../database/models/request';

/**
   * @param {object} options query option
   * @param {object} Request model /table
   * @returns {Promise} Promise resolved or rejected
   * @description get all rows where option matches
   */
  findRequest(options) {
    return Request.findAll({ where: { [Op.or]: Object.values(options) } });
  },
};

import DbServices from '../services/dbServices';
const { findRequest } = DbServices;

/**
 * search requested trip and approval trip controller
 * @param {Object} req - server request
 * @param {Object} res - server response
 * @returns {Object} - custom response
 */
const searchRequest = async (req, res) => {
  try {
    const {
      id, type, originCity, destinationCity, departureDate, reason, accommodation, approvalStatus
    } = req.params;

    if (id || type || originCity || destinationCity || departureDate || reason || accommodation
      || approvalStatus || approvalStatus) {
      const boolApprovalStatus = JSON.parse(approvalStatus);

      const options = {
        id,
        type,
        originCity,
        destinationCity,
        departureDate,
        reason,
        accommodation,
        boolApprovalStatus
      };

      const searchResult = await findRequest(options);
      if (searchResult.length <= 0) return response(res, 404, 'error', { message: messages.noResult });
      return response(res, 201, 'success', searchResult);
    }
  } catch (error) {
    return response(res, 400, 'error', { message: messages.error });
  }
};

export default {
  requestTrip,
  searchRequest
};

Answer 1

解决方法很简单:

//converting properties of object into an array of small objects
    const arrayOptions = Object.entries(options).map(([k, v]) => ({ [k]: v }));

//findRequest function       
 findRequest(options) {
            return Request.findAll({ where: { [Op.or]: arrayOptions } });
          },