Javascript 如何防止续集添加'；Id'；要使用列名吗？

我发现当我使用Sequelize“Id”进行查询时，会将其添加到我的列名末尾，但我不确定如何指示Sequelize不要这样做

我为Sequelize创建了一个实体数据模型，如下所示：

function initializeDataModel(sequelize) {
    var dataModel = { };    

    dataModel.Playlist = this.sequelize.define('playlist', {
        name: Sequelize.STRING,
    }); 

    dataModel.PlaylistEntry = this.sequelize.define('playlist_entry', {
        playlist: {
            name: 'playlist',
            type: Sequelize.INTEGER,
            references: {
                // This is a reference to another model
                model: dataModel.Playlist,

                // This is the column name of the referenced model
                key: 'id'

                // This declares when to check the foreign key constraint. PostgreSQL only.
                //deferrable: Sequelize.Deferrable.INITIALLY_IMMEDIATE 
            }               
        },
        track: Sequelize.INTEGER
    });                

    dataModel.PlaylistEntry.belongsTo(
         dataModel.Playlist,
         { as: 'Playlist', foreignKey: { name: 'playlist' }});

    dataModel.Playlist.hasMany(dataModel.PlaylistEntry);

    return dataModel;
}    

    eagerIncludes.push(this.dataModel.PlaylistEntry);

    this.dataModel.Playlist.find({
       where: { id: playlistId },
       limit: limit,
       offset: offset,
       include: eagerIncludes                 
    }).then(function (results) {
        callback(results, options, undefined);
    }).catch(function (error) {
        callback(undefined, options, error);
    });

“playlist_entry”（在MariaDB中）表中的字段如下所示：

function initializeDataModel(sequelize) {
    var dataModel = { };    

    dataModel.Playlist = this.sequelize.define('playlist', {
        name: Sequelize.STRING,
    }); 

    dataModel.PlaylistEntry = this.sequelize.define('playlist_entry', {
        playlist: {
            name: 'playlist',
            type: Sequelize.INTEGER,
            references: {
                // This is a reference to another model
                model: dataModel.Playlist,

                // This is the column name of the referenced model
                key: 'id'

                // This declares when to check the foreign key constraint. PostgreSQL only.
                //deferrable: Sequelize.Deferrable.INITIALLY_IMMEDIATE 
            }               
        },
        track: Sequelize.INTEGER
    });                

    dataModel.PlaylistEntry.belongsTo(
         dataModel.Playlist,
         { as: 'Playlist', foreignKey: { name: 'playlist' }});

    dataModel.Playlist.hasMany(dataModel.PlaylistEntry);

    return dataModel;
}    

    eagerIncludes.push(this.dataModel.PlaylistEntry);

    this.dataModel.Playlist.find({
       where: { id: playlistId },
       limit: limit,
       offset: offset,
       include: eagerIncludes                 
    }).then(function (results) {
        callback(results, options, undefined);
    }).catch(function (error) {
        callback(undefined, options, error);
    });

• id:INT（11）
• 播放列表：INT（11）
• 曲目：国际（11）

我正在执行的查询如下：

function initializeDataModel(sequelize) {
    var dataModel = { };    

    dataModel.Playlist = this.sequelize.define('playlist', {
        name: Sequelize.STRING,
    }); 

    dataModel.PlaylistEntry = this.sequelize.define('playlist_entry', {
        playlist: {
            name: 'playlist',
            type: Sequelize.INTEGER,
            references: {
                // This is a reference to another model
                model: dataModel.Playlist,

                // This is the column name of the referenced model
                key: 'id'

                // This declares when to check the foreign key constraint. PostgreSQL only.
                //deferrable: Sequelize.Deferrable.INITIALLY_IMMEDIATE 
            }               
        },
        track: Sequelize.INTEGER
    });                

    dataModel.PlaylistEntry.belongsTo(
         dataModel.Playlist,
         { as: 'Playlist', foreignKey: { name: 'playlist' }});

    dataModel.Playlist.hasMany(dataModel.PlaylistEntry);

    return dataModel;
}    

    eagerIncludes.push(this.dataModel.PlaylistEntry);

    this.dataModel.Playlist.find({
       where: { id: playlistId },
       limit: limit,
       offset: offset,
       include: eagerIncludes                 
    }).then(function (results) {
        callback(results, options, undefined);
    }).catch(function (error) {
        callback(undefined, options, error);
    });

这导致：

ER_BAD_FIELD_ERROR: Unknown column 'playlist_entries.playlistId' in 'field list'

如有任何建议，将不胜感激。更改数据库中的列名不是一个选项

---

注意，尝试使用“”时会出现此问题。

根据一些实验，修复方法是将“foreignKey”属性添加到hasMany定义中：

entities.Playlist.hasMany(entities.PlaylistEntry, { foreignKey: 'playlist' });

---

值得一提的是，Sequelize添加了

Id

，以区分作为记录的关系属性和关系键。我不确定您是否可以使用相同的名称而不引起严重问题。如果我能够将“playlist\u id”字段名更改为“playlist\u entry”表中的“playlist\u id”，我应该进行哪些更改才能使其正常工作？通常，最好遵守Sequelize规定的约定，除非您无法更改架构。如果需要，可以使用选项调整默认值，但有一件事您不能做，那就是让属性和它存储的列使用相同的名称。我正在使用一个现有的数据库，并且有一些约定，例如字段和表名称都是小写的，并且是用下划线分隔的单词。我不确定我是否理解您所说的“您不能做的一件事是使用相同的名称存储属性和列”。这在Sequelize中是如何表现的？有什么例子吗？我的意思是不能有一个名为

x

的列和一个名为

x

的属性，因为这两个属性会发生冲突。这就是为什么它通常是

xId

或

x_id

，这取决于您是否愿意区分它们的键和任何已实例化的相关记录。