Question

我正在寻找容量大于或等于输入容量的房间。我正在使用$gte运算符，它总是返回一个空数组。

我正在使用sequelize和MySQL，当我尝试时，

let allRooms = await Room.findAll({
        where: {
          capacity: 9
        }
      });

I get an array containing the data I'm trying to get.

[ Room {
    dataValues:
     { id: 2,
       name: 'meeting room 1',
       amenities: <Buffer 00 00 00>,
       capacity: 9,
       createdAt: 2019-08-26T18:21:07.000Z,
       updatedAt: 2019-08-26T18:21:07.000Z,
       locationId: 1 }
   }
]

但是当我尝试

let allRooms = await Room.findAll({
        where: {
          capacity: { $gte: 5 }
        }
      });

it returns  // []

在我的终端上，我得到了

Executing (default): SELECT `id`, `name`, `amenities`, `capacity`, 
`createdAt`, `updatedAt`, `locationId` FROM `Rooms` AS `Room`
 WHERE `Room`.`capacity` = '[object Object]';

Answer 1

您尝试过

const Op = Sequelize.Op

let allRooms = await Room.findAll({
        where: {
          capacity: { [Op.gte]: 5 }
        }
      });

编辑：

别名必须在使用前声明

const Op = Sequelize.Op
const operatorsAliases = {
  $gte: Op.gte
}

let allRooms = await Room.findAll({
        where: {
          capacity: { $gte: 5 }
        }
      });

$ gte总是返回一个空数组

1 个答案: