Can't push array in function [duplicate]
up vote
-1
down vote
favorite
This question already has an answer here:
Using async/await with a forEach loop
12 answers
How do I return the response from an asynchronous call?
33 answers
I know this has been asked before but I can't seem to find the answer, my array can't push to datahasil, how to push array hasil to datahasil...??
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});
javascript node.js express mongoose
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
marked as duplicate by Grégory NEUT, CertainPerformance javascript
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Nov 9 at 8:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
-1
down vote
favorite
This question already has an answer here:
Using async/await with a forEach loop
12 answers
How do I return the response from an asynchronous call?
33 answers
I know this has been asked before but I can't seem to find the answer, my array can't push to datahasil, how to push array hasil to datahasil...??
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});
javascript node.js express mongoose
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
marked as duplicate by Grégory NEUT, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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Nov 9 at 8:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
theres no return value
– nyx97
Nov 9 at 8:27
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
Use a debugger...
– James
Nov 9 at 8:31
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This question already has an answer here:
Using async/await with a forEach loop
12 answers
How do I return the response from an asynchronous call?
33 answers
I know this has been asked before but I can't seem to find the answer, my array can't push to datahasil, how to push array hasil to datahasil...??
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});
javascript node.js express mongoose
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
This question already has an answer here:
Using async/await with a forEach loop
12 answers
How do I return the response from an asynchronous call?
33 answers
I know this has been asked before but I can't seem to find the answer, my array can't push to datahasil, how to push array hasil to datahasil...??
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});
This question already has an answer here:
Using async/await with a forEach loop
12 answers
How do I return the response from an asynchronous call?
33 answers
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], function(err, result) {
var datahasil =
result.forEach(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
}, )
datahasil.push(hasil)
})
});javascript node.js express mongoose
javascript node.js express mongoose
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
edited Nov 9 at 8:26
Ayush Gupta
3,0011642
3,0011642
asked Nov 9 at 8:23
Ahmad Sayadi
376
asked Nov 9 at 8:23
Ahmad Sayadi
376
asked Nov 9 at 8:23
Ahmad Sayadi
376
376
marked as duplicate by Grégory NEUT, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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Nov 9 at 8:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Grégory NEUT, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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Nov 9 at 8:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
theres no return value
– nyx97
Nov 9 at 8:27
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
Use a debugger...
– James
Nov 9 at 8:31
add a comment |
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
theres no return value
– nyx97
Nov 9 at 8:27
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
Use a debugger...
– James
Nov 9 at 8:31
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
theres no return value
– nyx97
Nov 9 at 8:27
theres no return value
– nyx97
Nov 9 at 8:27
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
Use a debugger...
– James
Nov 9 at 8:31
Use a debugger...
– James
Nov 9 at 8:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
You gave forEach function an asynchronous method removing its synchronous nature so instead, use map and await Promise.all on the resulting array:
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});answered Nov 9 at 8:29
dotconnor
1,037120
Rather thanpushing to an external array, probably better to just use the results of thePromise.all(one less variable, more pure)
– CertainPerformance
Nov 9 at 8:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You gave forEach function an asynchronous method removing its synchronous nature so instead, use map and await Promise.all on the resulting array:
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});answered Nov 9 at 8:29
dotconnor
1,037120
Rather thanpushing to an external array, probably better to just use the results of thePromise.all(one less variable, more pure)
– CertainPerformance
Nov 9 at 8:31
add a comment |
up vote
0
down vote
accepted
You gave forEach function an asynchronous method removing its synchronous nature so instead, use map and await Promise.all on the resulting array:
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});answered Nov 9 at 8:29
dotconnor
1,037120
Rather thanpushing to an external array, probably better to just use the results of thePromise.all(one less variable, more pure)
– CertainPerformance
Nov 9 at 8:31
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You gave forEach function an asynchronous method removing its synchronous nature so instead, use map and await Promise.all on the resulting array:
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});answered Nov 9 at 8:29
dotconnor
1,037120
You gave forEach function an asynchronous method removing its synchronous nature so instead, use map and await Promise.all on the resulting array:
const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});const data1 = await JadwalBooking.aggregate([{
$project: {
"keterangan": "$keterangan",
"jumlah_tersisa": "$datalayout.jumlah_kursi_tersisa",
"flag": "$flag",
"date1": "$datamisa.date",
}
},
{
$match: {
flag: 1
},
},
], async function(err, result) {
const datahasil = await Promise.all(result.map(async(item) => {
const hasil = await DetailBookingMisa.find({
$or: [{
flag: 3
},
{
flag: 4
}
],
$and: [{
jadwalbookingmisa: item._id
}]
})
return hasil;
});
console.log(datahasil);
});answered Nov 9 at 8:29
dotconnor
1,037120
edited Nov 9 at 8:32
answered Nov 9 at 8:29
dotconnor
1,037120
answered Nov 9 at 8:29
dotconnor
1,037120
answered Nov 9 at 8:29
dotconnor
1,037120
1,037120
Rather thanpushing to an external array, probably better to just use the results of thePromise.all(one less variable, more pure)
– CertainPerformance
Nov 9 at 8:31
add a comment |
Rather thanpushing to an external array, probably better to just use the results of thePromise.all(one less variable, more pure)
– CertainPerformance
Nov 9 at 8:31
Rather than
pushing to an external array, probably better to just use the results of the Promise.all (one less variable, more pure)– CertainPerformance
Nov 9 at 8:31
Rather than
pushing to an external array, probably better to just use the results of the Promise.all (one less variable, more pure)– CertainPerformance
Nov 9 at 8:31
add a comment |
suggestion: consider not to query database in a loop.
– scetiner
Nov 9 at 8:26
theres no return value
– nyx97
Nov 9 at 8:27
forEach + async is virtually never a good combination
– Bravo
Nov 9 at 8:29
Use a debugger...
– James
Nov 9 at 8:31