Sequelize MySQL從同一個表中選擇（左連接）

如何從同一個表中選擇兩次Sequelize？這裏是MySQL代碼：Sequelize MySQL從同一個表中選擇（左連接）

select b.*, a.parent_id 
from theSameTable as b left join theSameTable as a on b.parent_id = a.id

這裏是我的MySQL表

這裏是我的Sequelize代碼

const db = await ec.sequelize.define(tableDb, { 
    id: { 
     type: ec.Sequelize.INTEGER.UNSIGNED, 
     primaryKey: true, 
     autoIncrement: true 
    }, 
    url: ec.Sequelize.STRING(511), 
    url_hash: ec.Sequelize.STRING(32), 
    name: ec.Sequelize.STRING(511), 
    full_name: ec.Sequelize.STRING(511), 
    parent_id: ec.Sequelize.INTEGER(11).UNSIGNED, 
    cnt: ec.Sequelize.STRING(255), 
    chk: ec.Sequelize.INTEGER(1) 
}, { 
    indexes: [{ 
     unique: true, 
     fields: ['url_hash'] 
    }] 
}); 

await ec.sequelize.sync();

來源

2017-10-18 big.data

你是怎麼找到谷歌上搜索有關自聯接的sequelize？ – philipxy

[使用Sequelize查詢自連接，包括相關記錄]的可能重複（https://stackoverflow.com/questions/40294776/query-self-join-with-sequelize-including-related-record） – philipxy

最後我找到了答案。也許對某人會有用。後

await ec.sequelize.sync();

寫

db.belongsTo(db, { 
    as: 'db2', 
    foreignKey: 'parent_id', 
    required: false 
}); 

let rows = await db.findAll({ 
    where: { 
     chk: 0 
    }, 
    include: [{ 
     model: db, 
     as: 'db2', 
     attributes: ['id', 'full_name'] 
    }], 
    raw: true, 
    limit: 20 
}).catch(function (err) { 
    console.log(err); 
});

來源

2017-10-19 13:12:49

Sequelize MySQL從同一個表中選擇（左連接）

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