javascript recursion to find children by array position

Question

I have a bunch of tasks and I need to find all children based on the task I click. I would like an array of position [ 1,2, 3] to come back so I can work with those rows.

Here is an example.

    var taskAR = [
    {
        "isParent": true,
        "parentID": null,
        "level": 0
    },
    {
        "isParent": true,
        "parentID": 0,
        "level": 1
    },
    {
        "isParent": true,
        "parentID": 1,
        "level": 2
    },
    {
        "isParent": false,
        "parentID": 2,
        "level": 3
    },
    {
        "isParent": false,
        "parentID": 2,
        "level": 3
    },
    {
        "isParent": false,
        "parentID": null,
        "level": 0
    },
    {
        "isParent": true,
        "parentID": null,
        "level": 0
    },
    {
        "isParent": false,
        "parentID": 7,
        "level": 1
    }

];


function getNestedChildren(arr, parentID) {
    var out = []
    for (var x = 0, len = arr.length; x < len ; x++) {
        var d = arr[x];

        if (d.parentID == parentID) {
            var children = getNestedChildren(arr, x);

            if (children.length) {
                d.parentID = x;
            }
            out.push(x)
        }
    }
    return out
    }

   getNestedChildren(this.master.tasks, parentID)

thanks for the help

And, do you have the ability to change the structure of your data? There are MUCH better ways to represent parent-cheild relationships than what you are using/ — talemyn
– talemyn, Commented Feb 20, 2015 at 15:38
Yep I can change it if there is a better way. The idea is to find all the children of the parent and then work with them both from the found array and on the display. — Killilea
– Killilea, Commented Feb 20, 2015 at 15:50
the problem I think is here: if (children.length) { d.parentID = x; } do I need to put another loop in there to get the position?? — Killilea
– Killilea, Commented Feb 20, 2015 at 15:52
BTW, a nice way to organize parent-child relations can be seen in DOM: a binary tree with left branch pointing to a firstChild and right branch pointing to a nextSibling. It might be augumented with the idea of skip list to eliminate the need of flattening... — Dmitry
– Dmitry, Commented Feb 20, 2015 at 16:21

Segfault · Accepted Answer · 2015-02-20 16:11:41Z

1

Two problems with this code: First is that you are not accumulating values. You need the output from the nested call to getNestedChildren to be returned in your out array. Second, I don't know why you are changing any parentIDs when this runs, it should be a read-only operation.

Take out

        if (children.length) {
            d.parentID = x;
        }

and replace it with

        out = out.concat(children);

answered Feb 20, 2015 at 16:11

Segfault

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