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Finding the position of super nested array of objects
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0
I'm trying to find the exact position and access all properties of the super nested array of objects.
I'm struggling to create a function where if I give index number as input parameter it should give me it's position in the array and also access all the properties in return.
Here is the sample array of object
I'm OK with ES6 and above solution too
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
Yeah I know this json object is scary enough to spend time and solve. Any kind of help is greatly appriciated.
for example if my function name is findChildIndex(22) it should return me something like this x.children[0].children[0].children[2]
Thank you!
javascript arrays reactjs object ecmascript-6
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
add a comment |
0
I'm trying to find the exact position and access all properties of the super nested array of objects.
I'm struggling to create a function where if I give index number as input parameter it should give me it's position in the array and also access all the properties in return.
Here is the sample array of object
I'm OK with ES6 and above solution too
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
Yeah I know this json object is scary enough to spend time and solve. Any kind of help is greatly appriciated.
for example if my function name is findChildIndex(22) it should return me something like this x.children[0].children[0].children[2]
Thank you!
javascript arrays reactjs object ecmascript-6
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
1
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47
add a comment |
0
0
0
I'm trying to find the exact position and access all properties of the super nested array of objects.
I'm struggling to create a function where if I give index number as input parameter it should give me it's position in the array and also access all the properties in return.
Here is the sample array of object
I'm OK with ES6 and above solution too
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
Yeah I know this json object is scary enough to spend time and solve. Any kind of help is greatly appriciated.
for example if my function name is findChildIndex(22) it should return me something like this x.children[0].children[0].children[2]
Thank you!
javascript arrays reactjs object ecmascript-6
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
I'm trying to find the exact position and access all properties of the super nested array of objects.
I'm struggling to create a function where if I give index number as input parameter it should give me it's position in the array and also access all the properties in return.
Here is the sample array of object
I'm OK with ES6 and above solution too
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
Yeah I know this json object is scary enough to spend time and solve. Any kind of help is greatly appriciated.
for example if my function name is findChildIndex(22) it should return me something like this x.children[0].children[0].children[2]
Thank you!
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
javascript arrays reactjs object ecmascript-6
javascript arrays reactjs object ecmascript-6
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
edited Mar 23 at 16:45
Harsh Makadia
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
asked Mar 23 at 16:37
Harsh MakadiaHarsh Makadia
1,31431228
1,31431228
1
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47
add a comment |
1
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47
1
1
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47
add a comment |
3 Answers
3
active
oldest
votes
3
You could recursively collect the indexes in the children arrays that lead to the target index:
function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));answered Mar 23 at 16:54
trincottrincot
134k1697134
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I formObj.children[0].children[0].children[2]from the array return [0,0,2] ?
– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable asvar properties = Obj.children[0].children[0].children[2]. If you can please help me
– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You meanvar properties = "Obj.children[0].children[0].children[2]"?
– trincot
Mar 23 at 19:17
|
show 7 more comments
2
You could use recursion and check if children of the element exists use for loop to iterate to through all the children and recursively apply the function of each child
const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
add a comment |
1
If i got your question correctly, You can do something like this:
const func=(obj,index, nested=0)=>
return Obj.index===index ? obj, nested : func(obj.children,index, nested+1)
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You could recursively collect the indexes in the children arrays that lead to the target index:
function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));answered Mar 23 at 16:54
trincottrincot
134k1697134
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I formObj.children[0].children[0].children[2]from the array return [0,0,2] ?
– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable asvar properties = Obj.children[0].children[0].children[2]. If you can please help me
– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You meanvar properties = "Obj.children[0].children[0].children[2]"?
– trincot
Mar 23 at 19:17
|
show 7 more comments
3
You could recursively collect the indexes in the children arrays that lead to the target index:
function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));answered Mar 23 at 16:54
trincottrincot
134k1697134
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I formObj.children[0].children[0].children[2]from the array return [0,0,2] ?
– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable asvar properties = Obj.children[0].children[0].children[2]. If you can please help me
– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You meanvar properties = "Obj.children[0].children[0].children[2]"?
– trincot
Mar 23 at 19:17
|
show 7 more comments
3
3
3
You could recursively collect the indexes in the children arrays that lead to the target index:
function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));answered Mar 23 at 16:54
trincottrincot
134k1697134
You could recursively collect the indexes in the children arrays that lead to the target index:
function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));function findIndexNested(data, index) []).findIndex(child =>
return result = findIndexNested(child, index)
);
if (result) return [i, ...result];
function findByPath(data, path)
for (let i of path) data = data.children[i];
return data
// Sample data
const data = "name": "branch 1","index": 1,"children": ["name": "sub child 1","index": 2,"children": ["name": "subx2 child 1","index": 3,"children": ["name": "subx3 child 1","index": 4,"children": ["name": "subx4 child 1","index": 21,"name": "subx4 child 2","index": 18],"name": "subx3 child 2","index": 6,"children": ["name": "subx4 child 1","index": 7,"name": "subx4 child 2","index": 21],"name": "subx3 child 3","index": 22]],"name": "sub child 2","index": 28]
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));answered Mar 23 at 16:54
trincottrincot
134k1697134
edited Mar 23 at 19:37
answered Mar 23 at 16:54
trincottrincot
134k1697134
answered Mar 23 at 16:54
trincottrincot
134k1697134
answered Mar 23 at 16:54
trincottrincot
134k1697134
134k1697134
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I formObj.children[0].children[0].children[2]from the array return [0,0,2] ?
– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable asvar properties = Obj.children[0].children[0].children[2]. If you can please help me
– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You meanvar properties = "Obj.children[0].children[0].children[2]"?
– trincot
Mar 23 at 19:17
|
show 7 more comments
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I formObj.children[0].children[0].children[2]from the array return [0,0,2] ?
– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable asvar properties = Obj.children[0].children[0].children[2]. If you can please help me
– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You meanvar properties = "Obj.children[0].children[0].children[2]"?
– trincot
Mar 23 at 19:17
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
Thanks, @trincot for the help! I really appreciate it.
– Harsh Makadia
Mar 23 at 16:57
how can I form
Obj.children[0].children[0].children[2] from the array return [0,0,2] ?– Harsh Makadia
Mar 23 at 18:32
how can I form
Obj.children[0].children[0].children[2] from the array return [0,0,2] ?– Harsh Makadia
Mar 23 at 18:32
You mean, as a string?
– trincot
Mar 23 at 18:52
You mean, as a string?
– trincot
Mar 23 at 18:52
yes like I want to store it in variable as
var properties = Obj.children[0].children[0].children[2]. If you can please help me– Harsh Makadia
Mar 23 at 19:14
yes like I want to store it in variable as
var properties = Obj.children[0].children[0].children[2]. If you can please help me– Harsh Makadia
Mar 23 at 19:14
Wait, that is not a string. You mean
var properties = "Obj.children[0].children[0].children[2]"?– trincot
Mar 23 at 19:17
Wait, that is not a string. You mean
var properties = "Obj.children[0].children[0].children[2]"?– trincot
Mar 23 at 19:17
|
show 7 more comments
2
You could use recursion and check if children of the element exists use for loop to iterate to through all the children and recursively apply the function of each child
const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
add a comment |
2
You could use recursion and check if children of the element exists use for loop to iterate to through all the children and recursively apply the function of each child
const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
add a comment |
2
2
2
You could use recursion and check if children of the element exists use for loop to iterate to through all the children and recursively apply the function of each child
const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
You could use recursion and check if children of the element exists use for loop to iterate to through all the children and recursively apply the function of each child
const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))const obj =
"name": "branch 1",
"index": 1,
"children": [
"name": "sub child 1",
"index": 2,
"children": [
"name": "subx2 child 1",
"index": 3,
"children": [
"name": "subx3 child 1",
"index": 4,
"children": [
"name": "subx4 child 1",
"index": 21
,
"name": "subx4 child 2",
"index": 18
]
,
"name": "subx3 child 2",
"index": 6,
"children": [
"name": "subx4 child 1",
"index": 7
,
"name": "subx4 child 2",
"index": 21
]
,
"name": "subx3 child 3",
"index": 22
]
]
,
"name": "sub child 2",
"index": 28
]
function find(obj,index)
if(obj.children)
for(let i = 0;i<obj.children.length;i++)
let x = find(obj.children[i],index);
if(x) return ...x,pos:i;
return obj.index === index ? obj : false;
console.log(find(obj,21))answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
answered Mar 23 at 16:44
Maheer AliMaheer Ali
17.2k21633
17.2k21633
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
add a comment |
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
Thanks for the help. Just one question here How can I know the index position where it is present like for 22 it could be Obj.children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:50
add a comment |
1
If i got your question correctly, You can do something like this:
const func=(obj,index, nested=0)=>
return Obj.index===index ? obj, nested : func(obj.children,index, nested+1)
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
add a comment |
1
If i got your question correctly, You can do something like this:
const func=(obj,index, nested=0)=>
return Obj.index===index ? obj, nested : func(obj.children,index, nested+1)
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
add a comment |
1
1
1
If i got your question correctly, You can do something like this:
const func=(obj,index, nested=0)=>
return Obj.index===index ? obj, nested : func(obj.children,index, nested+1)
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
If i got your question correctly, You can do something like this:
const func=(obj,index, nested=0)=>
return Obj.index===index ? obj, nested : func(obj.children,index, nested+1)
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
answered Mar 23 at 16:46
richard nelsonrichard nelson
11816
11816
add a comment |
add a comment |
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1
"it should give me it's position in the array and also access all the properties in return": please provide a concrete example of input and expected output.
– trincot
Mar 23 at 16:39
For example, my function name is findChildIndex(indexNumber) for example if 22 it should return me it's the position return me this children[0].children[0].children[2]
– Harsh Makadia
Mar 23 at 16:44
I think you need to ask yourself why this data is structured like this. Flat data is better for json.
– Joe Lloyd
Mar 23 at 16:44
@JoeLloyd This data is used for npmjs.com/package/react-d3-tree so can't help! :(
– Harsh Makadia
Mar 23 at 16:47