Question

我有两个belongsToMany模型：

const apptsModel = db.define('Appts', {
    id: {type: Sequelize.INTEGER, primaryKey: true, autoIncrement: true},
    [.....]
});

const UserDataModel = db.define('UserData', {
    id: {type: Sequelize.STRING, primaryKey: true},
    gender: {type: Sequelize.STRING},
    name_title: {type: Sequelize.STRING},
    name_first: {type: Sequelize.STRING},
    name_last: {type: Sequelize.STRING},
    [.....]
});

apptsModel.belongsToMany(UserDataModel, {through: 'ApptsToUsers'});
UserDataModel.belongsToMany(apptsModel, {through: 'ApptsToUsers'});

我想进行搜索：

1）查找至少有一个关联用户具有特定用户ID的所有约会。

2）返回该约会的所有关联用户。

我有工作的续集代码（1）：

var ret = connectors.Appts.findAll({
    include: [connectors.UserData],
    where: {'$UserData.id$': args.originatingUserID}
}).then((res) => res.map((item) => item.dataValues));

...但它只返回指定用户的关联用户数据。如何为每个约会返回所有关联用户的数据？

Answer 1

似乎还没有关于如何做到这一点的大量文档。这是有效的，所以我将在此处发布以供参考。

    getAllApptsForCurrentUser(_, args) {
       return Promise.resolve()
            .then(() => {
                //find all appointments and find those for which at least one 
                //participating user is the one specified in originatingUserID
                var appts = connectors.Appts.findAll({
                    include: [{
                        model: connectors.UserData,
                        where: {id: args.originatingUserID}
                    }],
                }).then((res) => res.map((item) => item.dataValues));
                return appts;
            })
            .then(appts => {
                //iterate returned appointments and perform a subquery on each,
                //finding the other participating users
                var arrayOfPromises = [];
                appts.forEach(function (appt) {
                    arrayOfPromises.push(
                        connectors.Appts.findOne({where: {id: appt.id}, order: [['apptDateTime']], include: [ connectors.UserData ] })
                    );
                });
                //Promise.all returns true when all promises passed to it have 
                //returned true, or when one of them returns false
                return Promise.all(arrayOfPromises);
            })
           .then(apptsWithJoinedData => {
               //console.log(apptsWithJoinedData);
               return apptsWithJoinedData;
           })
           .catch((err)=> {
                console.log(err);
            });
    }

如果有更好的方法，请告诉我。

Sequelize：查找基于相关属性的记录到多个记录 - 但仍然返回所有相关记录？

1 个答案: