Array of objects , group by parent's object
0
I've got an array of objects. I want to be able to group objects in parent's object , parent object is determined by broker: true. Is there a way to convert this:
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
Into something like this:
const data = [
{ id: 1, broker: true, chunks: [
{ id: 2, broker: false },
{ id: 3, broker: false },
]},
{ id: 4, broker: true, chunks: [
{ id: 5, broker: false },
]},
{ id: 6, broker: true, chunks: [
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
]},
];
javascript arrays sorting ecmascript-6
edited Nov 23 '18 at 18:57
dince12
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
|
show 1 more comment
0
I've got an array of objects. I want to be able to group objects in parent's object , parent object is determined by broker: true. Is there a way to convert this:
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
Into something like this:
const data = [
{ id: 1, broker: true, chunks: [
{ id: 2, broker: false },
{ id: 3, broker: false },
]},
{ id: 4, broker: true, chunks: [
{ id: 5, broker: false },
]},
{ id: 6, broker: true, chunks: [
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
]},
];
javascript arrays sorting ecmascript-6
edited Nov 23 '18 at 18:57
dince12
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
1
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09
|
show 1 more comment
0
0
0
I've got an array of objects. I want to be able to group objects in parent's object , parent object is determined by broker: true. Is there a way to convert this:
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
Into something like this:
const data = [
{ id: 1, broker: true, chunks: [
{ id: 2, broker: false },
{ id: 3, broker: false },
]},
{ id: 4, broker: true, chunks: [
{ id: 5, broker: false },
]},
{ id: 6, broker: true, chunks: [
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
]},
];
javascript arrays sorting ecmascript-6
edited Nov 23 '18 at 18:57
dince12
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
I've got an array of objects. I want to be able to group objects in parent's object , parent object is determined by broker: true. Is there a way to convert this:
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
Into something like this:
const data = [
{ id: 1, broker: true, chunks: [
{ id: 2, broker: false },
{ id: 3, broker: false },
]},
{ id: 4, broker: true, chunks: [
{ id: 5, broker: false },
]},
{ id: 6, broker: true, chunks: [
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
]},
];
javascript arrays sorting ecmascript-6
javascript arrays sorting ecmascript-6
edited Nov 23 '18 at 18:57
dince12
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
edited Nov 23 '18 at 18:57
dince12
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
edited Nov 23 '18 at 18:57
dince12
390215
edited Nov 23 '18 at 18:57
dince12
390215
edited Nov 23 '18 at 18:57
dince12
390215
390215
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
asked Nov 23 '18 at 18:51
Dorin MusteațaDorin Musteața
35110
35110
1
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09
|
show 1 more comment
1
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09
1
1
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09
|
show 1 more comment
4 Answers
4
active
oldest
votes
1
You could check the propery for broker and push either a new object to the result set, or push the object to the previous object's chunks array.
const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
add a comment |
0
Use reduce to add to an accumulator every time it runs into a true value, and if it runs into a false value, add to the previous object via the property chunks
data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]
["chunks"] || (a[len]["chunks"] = ), a[len]
["chunks"].push(cv), a);
}, );
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
add a comment |
-1
You have to go through the array, create a new one and evaluate the condition you need, if it is met, put it in the new array
Object.keys(data).forEach((value, index) => {
// code
})answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
Object.keys()for an array? O.o
– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like theObjectinObject.keysalready suggests, is this method for objects and not for arrays.
– Andreas
Nov 23 '18 at 19:19
add a comment |
-1
You are going through the keys of each object
Object.keys(data).forEach(key => {
console.log(data[index].property)
})found
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You could check the propery for broker and push either a new object to the result set, or push the object to the previous object's chunks array.
const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
add a comment |
1
You could check the propery for broker and push either a new object to the result set, or push the object to the previous object's chunks array.
const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
add a comment |
1
1
1
You could check the propery for broker and push either a new object to the result set, or push the object to the previous object's chunks array.
const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
You could check the propery for broker and push either a new object to the result set, or push the object to the previous object's chunks array.
const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }const
data = [{ id: 1, broker: true }, { id: 2, broker: false }, { id: 3, broker: false }, { id: 4, broker: true }, { id: 5, broker: false }, { id: 6, broker: true }, { id: 7, broker: false }, { id: 8, broker: false }, { id: 9, broker: false }],
grouped = data.reduce((r, o) => {
if (o.broker) {
r.push(Object.assign({}, o, { chunks: }));
} else {
r[r.length - 1].chunks.push(o);
}
return r;
}, );
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
answered Nov 23 '18 at 18:59
Nina ScholzNina Scholz
177k1391156
177k1391156
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
add a comment |
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
That will work out. Thanks.
– Dorin Musteața
Nov 23 '18 at 19:07
add a comment |
0
Use reduce to add to an accumulator every time it runs into a true value, and if it runs into a false value, add to the previous object via the property chunks
data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]
["chunks"] || (a[len]["chunks"] = ), a[len]
["chunks"].push(cv), a);
}, );
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
add a comment |
0
Use reduce to add to an accumulator every time it runs into a true value, and if it runs into a false value, add to the previous object via the property chunks
data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]
["chunks"] || (a[len]["chunks"] = ), a[len]
["chunks"].push(cv), a);
}, );
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
add a comment |
0
0
0
Use reduce to add to an accumulator every time it runs into a true value, and if it runs into a false value, add to the previous object via the property chunks
data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]
["chunks"] || (a[len]["chunks"] = ), a[len]
["chunks"].push(cv), a);
}, );
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
Use reduce to add to an accumulator every time it runs into a true value, and if it runs into a false value, add to the previous object via the property chunks
data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]
["chunks"] || (a[len]["chunks"] = ), a[len]
["chunks"].push(cv), a);
}, );
const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);const data = [
{ id: 1, broker: true },
{ id: 2, broker: false },
{ id: 3, broker: false },
{ id: 4, broker: true },
{ id: 5, broker: false },
{ id: 6, broker: true },
{ id: 7, broker: false },
{ id: 8, broker: false },
{ id: 9, broker: false },
];
let result = data.reduce((a, cv) => {
let len = a.length - 1;
return (cv["broker"]) ? (a.push(cv), a) : (a[len]["chunks"] || (a[len]["chunks"] = ), a[len]["chunks"].push(cv), a);
}, );
console.log(result);answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
answered Nov 23 '18 at 19:06
zfrischzfrisch
4,42811024
4,42811024
add a comment |
add a comment |
-1
You have to go through the array, create a new one and evaluate the condition you need, if it is met, put it in the new array
Object.keys(data).forEach((value, index) => {
// code
})answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
Object.keys()for an array? O.o
– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like theObjectinObject.keysalready suggests, is this method for objects and not for arrays.
– Andreas
Nov 23 '18 at 19:19
add a comment |
-1
You have to go through the array, create a new one and evaluate the condition you need, if it is met, put it in the new array
Object.keys(data).forEach((value, index) => {
// code
})answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
Object.keys()for an array? O.o
– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like theObjectinObject.keysalready suggests, is this method for objects and not for arrays.
– Andreas
Nov 23 '18 at 19:19
add a comment |
-1
-1
-1
You have to go through the array, create a new one and evaluate the condition you need, if it is met, put it in the new array
Object.keys(data).forEach((value, index) => {
// code
})answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
You have to go through the array, create a new one and evaluate the condition you need, if it is met, put it in the new array
Object.keys(data).forEach((value, index) => {
// code
})Object.keys(data).forEach((value, index) => {
// code
})Object.keys(data).forEach((value, index) => {
// code
})answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 18:59
Darwin RodríguezDarwin Rodríguez
1
1
Object.keys()for an array? O.o
– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like theObjectinObject.keysalready suggests, is this method for objects and not for arrays.
– Andreas
Nov 23 '18 at 19:19
add a comment |
Object.keys()for an array? O.o
– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like theObjectinObject.keysalready suggests, is this method for objects and not for arrays.
– Andreas
Nov 23 '18 at 19:19
Object.keys() for an array? O.o– Andreas
Nov 23 '18 at 18:59
Object.keys() for an array? O.o– Andreas
Nov 23 '18 at 18:59
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Object.keys(data).forEach(key => { console.log(data[key].property) })
– Darwin Rodríguez
Nov 23 '18 at 19:16
Like the
Object in Object.keys already suggests, is this method for objects and not for arrays.– Andreas
Nov 23 '18 at 19:19
Like the
Object in Object.keys already suggests, is this method for objects and not for arrays.– Andreas
Nov 23 '18 at 19:19
add a comment |
-1
You are going through the keys of each object
Object.keys(data).forEach(key => {
console.log(data[index].property)
})found
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
add a comment |
-1
You are going through the keys of each object
Object.keys(data).forEach(key => {
console.log(data[index].property)
})found
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
add a comment |
-1
-1
-1
You are going through the keys of each object
Object.keys(data).forEach(key => {
console.log(data[index].property)
})found
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
You are going through the keys of each object
Object.keys(data).forEach(key => {
console.log(data[index].property)
})found
Object.keys(data).forEach(key => {
console.log(data[index].property)
})Object.keys(data).forEach(key => {
console.log(data[index].property)
})answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
answered Nov 23 '18 at 19:13
Darwin RodríguezDarwin Rodríguez
1
1
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
add a comment |
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Why the same "answer" twice?
– Andreas
Nov 23 '18 at 19:31
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
Remember that it is an array of objects therefore you can use foreach
– Darwin Rodríguez
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
See my comment on your other answer...
– Andreas
Nov 23 '18 at 19:32
add a comment |
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1
where is you specific problem?
– Nina Scholz
Nov 23 '18 at 18:55
@NinaScholz what do you mean? I want to be able to group objects in parent's object. I cant achieve that kind of grouping.
– Dorin Musteața
Nov 23 '18 at 18:58
have you tried anything?
– Nina Scholz
Nov 23 '18 at 18:59
@NinaScholz Of course , that's why asked , thanks for the right answer anyway.
– Dorin Musteața
Nov 23 '18 at 19:07
"Of course , that's why asked" - Then why didn't you post your code? Just copy&pasting the answer from somebody won't help you understand what you've done wrong.
– Andreas
Nov 23 '18 at 19:09