Question

我正在使用sequelize事务（托管），当错误引发回滚并不是真的发生时，也许我在这里做错了什么？

我的测试用例：当代码块2抛出错误时，代码块1中完成的事情不会回滚，请帮忙。

return sequelize.transaction(function (t) {
        return db.VenueTag.destroy({where: {venueId: venue.venueId}}, {transaction: t}).then(function () {
        }, {transaction: t}).then(function () {
        //block 1
            return db.VenueTag.upsert({
            ...
            });
        }, {transaction: t}).then(function () {
        //block 2
            //update venue details
            return venue.updateAttributes({
                ...
            },{ transaction: t});
        });
    }).then(function (result) {
        // Transaction has been committed
        // result is whatever the result of the promise chain returned to the transaction callback
        return res.status(200).json(result);
    }).catch(function (err) {
        // Transaction has been rolled back
        // err is whatever rejected the promise chain returned to the transaction callback
        return res.status(500).send({
            message: errorHandler.getErrorMessage(err)
        });
    });

Answer 1

你的承诺链中有错误，试试这个：

return sequelize.transaction(function (t) {
    return db.VenueTag.destroy({where: {venueId: venue.venueId}}, {transaction: t})
    .then(function () {
    //block 1
        return db.VenueTag.upsert({
        ...
        }, {transaction: t});
    .then(function () {
    //block 2
        //update venue details
        return venue.updateAttributes({
            ...
        },{ transaction: t});
    });
}).then(function (result) {
    // Transaction has been committed
    // result is whatever the result of the promise chain returned to the transaction callback
    return res.status(200).json(result);
}).catch(function (err) {
    // Transaction has been rolled back
    // err is whatever rejected the promise chain returned to the transaction callback
    return res.status(500).send({
        message: errorHandler.getErrorMessage(err)
    });
});

Answer 2

如果您要迎合ES7受众或使用node.js。您可以使用async / await方法来简化回调/承诺链：

/*Global Setup*/

async function deleteVenueTag () {
  try {
    let transaction = await sequelize.transaction();
    let VenueTag = await db.findOne({where: {venueId: venue.venueId}}, {transaction});
    await VenueTag.destroy({ transaction });
    await VenueTag.upsert({ transaction });
    await venue.updateAttributes({ transaction });
    await transaction.commit();
  }
} catch (error) {
  transaction.rollback();
}

Answer 3

如果你在使用下面这样的东西

return Users.update(updateUser, {
    where: {
        uid: sessionUser.uid,
        status: 'ACTIVE'
    },
    transaction: t     //second parameter is "options", so transaction must be in it
})

然后 transaction 是第二个选项参数中的一个键，而不是第三个

Sequelize事务错误抛出时不执行回滚

3 个答案: