如何使用别名选择列

Question

如何执行这样的 sql 查询？

SELECT column_name AS alias_name FROM table_name;

示例：我希望将“first”列选为“firstname”

Table.findAll({
      attributes: [id,"first"]
    })
    .then(function(posts) {
        res.json(posts);
    })

Answer 1

Table.findAll({
  attributes: ['id', ['first', 'firstName']] //id, first AS firstName
})
.then(function(posts) {
  res.json(posts);
});

Answer 2

Sequelize 也支持直接在模型定义上定义列名。

Sequelize Docs他们提到关于列定义的field属性。

例如：（取自 Docs 本身）

const { Model, DataTypes, Deferrable } = require("sequelize");

class Foo extends Model { }
Foo.init({
    // You can specify a custom column name via the 'field' attribute:
    fieldWithUnderscores: {
        type: DataTypes.STRING, 
        field: 'field_with_underscores'
    },
}, {
    sequelize,
    modelName: 'foo'
});

感谢这个答案

Answer 3

您需要使用row.get('newname')访问由attributes别名的列

由于某种原因，只执行row.newname或row.oldname将无法像处理非别名名称那样工作：

• https://github.com/sequelize/sequelize/issues/10592

• https://sequelize.org/v5/manual/models-usage.html记录它：

   Project.findOne({ where: {title: 'aProject'}, attributes: ['id', ['name', 'title']] }).then(project => { // project will be the first entry of the Projects table with the title // 'aProject' || null // project.get('title') will contain the name of the project })

  但是https://sequelize.org/v5/manual/models-usage.html没有提到它，这令人困惑：

   instance.field // is the same as instance.get('field')