score:18

Accepted answer

Are you using D3.js as your tag implies? Because in that case, you can just use d3.keys():

var data = [{"A":20,"B":32,"C":27,"D":30,"E":40}];
d3.keys(data[0]); // ["A", "B", "C", "D", "E"] 

If you want the sum of all the values, you might be better off using d3.values() and d3.sum():

var data = [{"A":20,"B":32,"C":27,"D":30,"E":40}, {"F":50}];
// get total of all object totals
var total = d3.sum(data, function(d) {
    // get total of a single object's values
    return d3.sum(d3.values(d));
});
total; // 199

score:0

A for-in-loop does the trick. On one object it looks like this:

var o = {
    a: 5,
    b: 3
};
var num = 0;

for (var key in o) {
    num += o[key];
}
alert(num);

score:0

Try this. It is simple:

var a = [{"A":20,"B":32,"C":27,"D":30,"E":40}];

for(var i in a){
  for(var j in a[i]){
    console.log(j); // shows key
    console.log(a[i][j]); // shows value
  }
}

score:0

I think this should be parsed recursively like below

var getKeys = function(previousKeys,obj){
        var currentKeys = Object.keys(obj);
        previousKeys = previousKeys.concat(currentKeys);
        for(var i=0;i<currentKeys.length;i++){
            var innerObj = obj[currentKeys[i]];
            if(innerObj!==null && typeof innerObj === 'object' && !Array.isArray(innerObj)){
            return this.getKeys(previousKeys,innerObj);
            }
        }
        return previousKeys;
    }

usage: getKeys([],{"a":"1",n:{c:"3",e:{ f:4,g:[1,2,3]}}})

Result: ["a", "n", "c", "e", "f", "g"]

score:0

var _ = require('underscore');

var obj = [{"A":20,"B":32,"C":27,"D":30,"E":40},{"F":50}, {"G":60,"H":70},{"I":80}];

var keys = [], values = [];



_.each(obj, function(d) {

     keys.push(_.keys(d));

     values.push(_.values(d));
});


// Keys   ->  [ 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I' ]
console.log('Keys   -> ', _.flatten(keys ));
// Values ->  [ 20, 32, 27, 30, 40, 50, 60, 70, 80 ]
console.log('Values -> ', _.flatten(values));

score:0

stop reinventing the wheel !

Object.keys()

MDN

score:2

Try using the JavaScript for..in statement:

var getKeys = function(arr) {
  var key, keys = [];
  for (i=0; i<arr.length; i++) {
    for (key in arr[i]) {
      keys.push(key);
    }
  }
  return keys;
};

var a = [{"A":20, "B":32, "C":27, "D":30, "E":40}, {"F":50}]
getKeys(a); // => ["A", "B", "C", "D", "E", "F"]

score:2

I think this is the simplest.

var a = [{"A":20,"B":32,"C":27,"D":30,"E":40}];
Object.keys( a[0] );

Result :

["A", "B", "C", "D", "E"]

score:4

Use for .. in:

var result = 0;

for (var i = jsonArray.length - 1; i >= 0; --i) {
    var o = jsonArray[i];
    for (var key in o) {
      if (o.hasOwnProperty(key)) {
        result += o[key];
      }
    }
    // in your code, you return result here, 
    // which might not give the right result 
    // if the array has more than 1 element
}

return result;

score:4

var easy = {
    a: 1,
    b: 2,
    c: 3
}

var keys = [], vals = []
for (var key in easy) {
    keys.push(key)
    vals.push(easy[key])
}

alert(keys+" - tha's how easy baby, it's gonna be")
alert(vals+" - tha's how easy baby, it's gonna be")

defensively

Including @Sahil's defensive method...

for (var key in easy) {
    if (easy.hasOwnProperty(key)) {
        keys.push(key)
        vals.push(easy[key])
    }
}

score:11

All of the current posted solutions have a problem. None of them check for object.hasOwnProperty(prop) while iterating over an object using a for...in loop. This might cause phantom keys to appear if properties are added to the prototype.

Quoting Douglas Crockford

Be aware that members that are added to the prototype of the object will be included in the enumeration. It is wise to program defensively by using the hasOwnProperty method to distinguish the true members of the object.

Adding a check for hasOwnProperty to maerics' excellent solution.

var getKeys = function (arr) {
        var key, keys = [];
        for (i = 0; i < arr.length; i++) {
            for (key in arr[i]) {
                if (arr[i].hasOwnProperty(key)) {
                    keys.push(key);
                }
            }
        }
        return keys;
    };

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