Node.js deleteMany仅返回更改流中删除的1个值
我有一个deleteMany请求,但在筛选deleteMany返回值的上下文时遇到了困难。它只返回从pusherjs中删除的1个值 这是我在服务器端的变更流代码和推送程序代码Node.js deleteMany仅返回更改流中删除的1个值,node.js,reactjs,mongodb,mongoose,pusher,Node.js,Reactjs,Mongodb,Mongoose,Pusher,我有一个deleteMany请求,但在筛选deleteMany返回值的上下文时遇到了困难。它只返回从pusherjs中删除的1个值 这是我在服务器端的变更流代码和推送程序代码 if (schedules.operationType === 'delete') { const scheduleDetails = schedules.documentKey; pusher.trigger('schedules', 'deleted', {
if (schedules.operationType === 'delete') {
const scheduleDetails = schedules.documentKey;
pusher.trigger('schedules', 'deleted', {
_id: scheduleDetails._id,
teamOne: scheduleDetails.teamOne,
teamTwo: scheduleDetails.teamTwo,
user: scheduleDetails.user,
isDone: scheduleDetails.isDone,
isStarted: scheduleDetails.isStarted,
date: scheduleDetails.date,
gameEvent: scheduleDetails.gameEvent,
});
}
这是我客户端的推送代码。顺便说一下,我正在用React。它存储在我的上下文api中
ScheduleChannel.bind('deleted', ({ deletedSchedule }) => {
console.log(deletedSchedule);
setScheduleList(
scheduleList.filter((schedule) => schedule._id !== deletedSchedule._id)
);
});
这是我要求的代码
exports.deleteallmatch = async (req, res) => {
try {
const { sub } = req.user;
const deletedMatches = await Schedule.deleteMany({ user: sub });
return res.status(201).json({
message: 'All of your schedule is successfully deleted!',
deletedMatches,
});
} catch (err) {
return res.status(400).json({
message: 'Something went wrong.',
});
}
};
删除请求很好,但我希望在我的应用程序中具有实时性。因为只发送了一个数据,而不是多个数据。我怎样才能解决这个问题
deleteMany()方法返回一个包含三个字段的对象:
let deleted_items = await Schedule.find({ user: sub });
await Schedule.deleteMany({ user: sub });
return res.status(201).json({
message: 'All of your schedule is successfully deleted!',
deleted_items,
});