sequelize.literal 返回原属性名称和重命名后的属性

Question

我想在执行join后重命名属性，我尝试在sequelize.literal中使用 sequelize.literal 但它返回原始名称和修改后的名称。 如何从结果中删除原点名称？

const result = await User.findAll({
  attributes: [
    'name',
    'id',
    [Sequelize.literal('Company.name'), 'companyName'],
    [Sequelize.literal('Employee.type'), 'employeeType'],
  ],
  include: [
    { model: Company, attributes: ['name'], required: false },
    { model: Employee, attributes: ['type'], required: true },,
  ],
  raw: true,
});

结果：

...
"companyName": "Test",
"employeeType": "Junior",
"Company.name": "Test",
"Employee.type": "Junior",

预期结果：

...
"companyName": "Test",
"employeeType": "Junior",

Answer 1

我刚刚找到了这个问题的解决方案，但我对其他想法持开放态度

这就是我所做的，它对我很有效。

const result = await User.findAll({
  attributes: [
    'name',
    'id',
    'Company.name',
    [Sequelize.literal('`Company`.`name`'), 'companyName'],
    'Employee.type',
    [Sequelize.literal('`Employee`.`type`'), 'employeeType'],
  ],
  include: [
    { model: Company, attributes: [], required: false },
    { model: Employee, attributes: [], required: true },,
  ],
  raw: true,
});