我有3个函数，我想逐步调用这个函数，例如当我调用第一个函数并获取结果时，我必须调用第二个函数并传递从第一次调用返回的参数。在完成第二个调用后，我必须调用第三个函数并传递从第二个函数返回的参数。
#1:
getCategory = function (branch_id) {
var deferred = q.defer();
var categoryData;
var query = 'SELECT id,name,price,currency FROM category where branch_id=?';
pool.getConnection(function (err, connection) {
connection.query(query, [branch_id], function (error, row, fields) {
if (error) {
deferred.reject(error);
} else {
connection.release();
deferred.resolve(row);
}
});
});
return deferred.promise; }
#2:
getRoom = function (branch_id, categoryID) {
var deferred = q.defer();
var roomData;
var roomSql = 'SELECT id,room_no,name,price,currency FROM room where branch_id=? and category_id=?';
pool.getConnection(function (err, connection) {
connection.query(roomSql, [branch_id, categoryID], function (error, row, fields) {
if (err) {
deferred.reject(err);
} else {
connection.release();
deferred.resolve(row);
}
});
});
return deferred.promise;
}
#3:
getReservationL = function (room_id, start_date, end_date) {
var deferred = q.defer();
var reservationData;
var reservationSql = 'SELECT d.id,d.create_date,d.update_date,d.room_id,d.status_id,d.start_date,d.end_date, ' +
' s.name as status_name,a.id as reservation_id,a.person_no as person_no, p.first_name,p.last_name,p.email ' +
' FROM reservation_detail d ' +
' inner join reservation_status s on d.status_id=s.id ' +
' inner join reservation a on d.reservation_id=a.id ' +
' inner join person p on a.person_no=p.personal_no ' +
' where d.room_id=? and d.start_date >? and d.start_date<?';
pool.getConnection(function (err, connection) {
connection.query(reservationSql, [room_id, start_date, end_date], function (error, row, fields) {
if (err) {
deferred.reject(err);
} else {
connection.release();
deferred.resolve(row);
}
});
});
return deferred.promise;
}
我需要这样的东西：
data = getCategory()
for(i=0;i<data.length;i++){
data[i].room = getRoom(data[i].id);
for(j=0;j<data[i].room[j].length;j++){
data[i].room[j].reservation = getReservationL(data[i].room[j].id);
}
}
如何在 NodeJS 中使用 Promise 或 Callback 来实现这个伪代码。我更喜欢使用承诺。
UPDATE#1第二次迭代后我有这样的数据
[
{
"id": 64,
"name": "VIP",
"price": 116.5,
"currency": "USD",
"room": [
{
"id": 44,
"room_no": "101",
"name": "A",
"price": 100,
"currency": "EUR"
},
{
"id": 274,
"room_no": "505",
"name": "a",
"price": 1,
"c\r\nurrency": "GEL"
}
]
},
{
"id": 74,
"name": "SUPER VIP",
"price": 110,
"currency": "EUR",
"room": [
{
"id": 54,
"room_no": "102",
"name": "A",
"price": 100,
"currency": "GEL"
},
{
"id": 284,
"room_no": "606",
"name": "a",
"price": 1,
"currency": "GEL"
}
]
},
{
"id": 84,
"name": "DOUBLE",
"price": 110,
"currency": "GEL",
"room": [
{
"id": 204,
"room_no": "103",
"name": "b",
"price": 120,
"currency": "GEL"
}
]
}
]
我想迭代每个房间的每个类别。
getCategory(branch_id).then(firstRecords => {
let promises = firstRecords.map(function (record) {
return getRoom(branch_id, record.id)
.then(roomData => {
var res = Object.assign({}, record, { room: roomData });
return res;
})
});
return Promise.all(promises);
//HERE i have data that is placed above.
}).then(secondRecords => {
let promises = secondRecords.map(function (category) {
return category.room;
}).map(function (rooms) {
console.log('SECOND', rooms);
return rooms;
}).map(function (reservation) {
console.log('THERD', reservation);
return reservation;
})
return Promise.all(promises);
}).then(reservation => {
console.log("Reservation", reservation);
})
UPDATE#2最终解决方案在这里
getCategory(branch_id) .then( categories => {
let roomPromises = categories.map( category => {
return getRoom(branch_id, category.id)
.then( rooms => Object.assign({}, category, { rooms }) ) });
return Promise.all(roomPromises) }) .then( category_rooms => {
let finalPromise = category_rooms.map( category => {
let reservationPromises = category.rooms.map( room => {
return getReservationL(room.id, start_date, end_date)
.then( reservations => Object.assign({}, room, { reservations }) )
})
return Promise.all(reservationPromises)
.then( room_reservations => {
return Object.assign({}, category, { rooms: room_reservations })
}); })
// const flattenPromise = finalPromise.reduce( (a, b) => a.concat(b), []); // return Promise.all(flattenPromise);
return Promise.all(finalPromise) }) .then( data => console.log('final: ', data) )
