搜索嵌套对象中特定键的值

Question

I have this nested object:

{
    "PINS" : {
        "2017" : {
            "Nov-2017" : {
                "VJkRWX7pTSl_5w1Np" : {
                    "pin" : "6K3jP5vLyN",
                    "say": "Hello"
                },
                "MsdsXiO9G9mwM3Qa" : {
                    "pin" : "hnPKh7ywvT",
                    "say": "Hello"
                }
            },
            "Dec-2017" : {
                "Mm35Gjb-nY0k2TV" : {
                    "pin" : "xWwaNNE2XG",
                    "say": "Hello" 
                },
                "WzajCLEJmJHmzg0" : {
                    "pin" : "vMU1mKbZAi",
                    "say": "Hello"
                }
            }
        },
        "2018" : {
            "Jan-2018" : {
                "Wf8E1unVaOh03a43" : {
                    "pin" : "qXJCQREATD",
                    "say": "Hello"
                },
                "JZqP8fVCLSja6J82v" : {
                    "pin" : "o5D8S8Lvtb",
                    "say": "Hello"
                }
            },
            "Feb-2018" : {
                "lMMAKNLy8jtnnXAN" : {
                    "pin" : "9zDuHcw6qH",
                    "say": "Hello" 
                },
                "e9EV3HDKCceM" : {
                    "pin" : "kPllwcoaob",
                    "say": "Hello" 
                }
            }
        }
    }
}

what I need is to find for all the 'pin' keys, and get their values, to put them into an array.

Exactly, I need to have an array like this:

['6K3jP5vLyN', 'hnPKh7ywvT', 'xWwaNNE2XG', 'vMU1mKbZAi', 'qXJCQREATD', 'o5D8S8Lvtb', '9zDuHcw6qH', 'kPllwcoaob']

I have tried:

const array = [];

function iter(obj){
    for(key in obj){
        if(obj.pin)
        array.push(obj.pin);
        if(obj[key]!==null && typeof obj[key]==="object"){
            iter(obj[key]);
        }
    }
}
iter(obj);

But I get the values of each key twice. Is there some improved way to do this?

Answer 1

You can use a recursive method to flatten the nested object to it's leaf values.

Option 1 - the value is the only property on the leaf (original answer before question updated)

The method extracts the current values to an array using Object.values() . It iterates the array with Array.map() , and flattens any value which is an object. The result of each run is spread into Array.concat() to flatten the nested arrays.

 const data = {"PINS":{"2017":{"Nov-2017":{"VJkRWX7pTSl_5w1Np":{"pin":"6K3jP5vLyN"},"MsdsXiO9G9mwM3Qa":{"pin":"hnPKh7ywvT"}},"Dec-2017":{"Mm35Gjb-nY0k2TV":{"pin":"xWwaNNE2XG"},"WzajCLEJmJHmzg0":{"pin":"vMU1mKbZAi"}}},"2018":{"Jan-2018":{"Wf8E1unVaOh03a43":{"pin":"qXJCQREATD"},"JZqP8fVCLSja6J82v":{"pin":"o5D8S8Lvtb"}},"Feb-2018":{"lMMAKNLy8jtnnXAN":{"pin":"9zDuHcw6qH"},"e9EV3HDKCceM":{"pin":"kPllwcoaob"}}}}}; const flattenObj = (obj) => [].concat(...Object.values(obj).map((o) => typeof o === 'object' ? flattenObj(o) : o)); const result = flattenObj(data); console.log(result); 

Option 2 - the value is not the only property on the leaf

If your data contains other keys, this variant uses Object.entries() to extract a specific key:

 const data = {"PINS":{"2017":{"Nov-2017":{"VJkRWX7pTSl_5w1Np":{"pin":"6K3jP5vLyN","say":"Hello"},"MsdsXiO9G9mwM3Qa":{"pin":"hnPKh7ywvT","say":"Hello"}},"Dec-2017":{"Mm35Gjb-nY0k2TV":{"pin":"xWwaNNE2XG","say":"Hello"},"WzajCLEJmJHmzg0":{"pin":"vMU1mKbZAi","say":"Hello"}}},"2018":{"Jan-2018":{"Wf8E1unVaOh03a43":{"pin":"qXJCQREATD","say":"Hello"},"JZqP8fVCLSja6J82v":{"pin":"o5D8S8Lvtb","say":"Hello"}},"Feb-2018":{"lMMAKNLy8jtnnXAN":{"pin":"9zDuHcw6qH","say":"Hello"},"e9EV3HDKCceM":{"pin":"kPllwcoaob","say":"Hello"}}}}}; const flattenObjKey = (obj, key) => [].concat(...Object.entries(obj) .map(([k, v]) => typeof v === 'object' ? flattenObjKey(v, key) : (k === key ? v : []) )); const result = flattenObjKey(data, 'pin'); console.log(result); 

Answer 2

It looks like your data is consistently structured, so this is fairly simple.

const pins = [];
for (let year in pins) {
    for (let month in year) {
        for (let key in months) {
            pins.push(key.pin);
        }
    }
}

If it's not consistently structured, you'll need to extract a recursive solution that looks for pin keys.