Can't convert Javascript array to object properly

Question

I have been struggling with this problem the entire day. I feel like it's super solvable, and I'm not sure where I'm going wrong. Each time I go to post this, I feel like I come up with a different solution that doesn't end up working.

I'm looking to do the following:

var someObj = {};

// @param key - string - in the form of "foo-bar-baz" or "foo-bar". 
// (i won't know the number of segments ahead of time.)
// @param value - string - standard, don't need to act on it

function buildObject( key, value ) {

  var allKeys = key.split("-");

  // do stuff here to build someObj

}

Essentially the key will always take the format of key-key-key where i want to build someObj[key1][key2][key3] = value.

This JSFiddle contains a longer example with a sample layout of the data structure I want to walk away with.

Thanks so much for any help you can give.

Mike Samuel · Accepted Answer · 2011-05-04 00:13:45Z

2

var someObj = {};

// @param key - string - in the form of "foo-bar-baz" or "foo-bar". 
// (i won't know the number of segments ahead of time.)
// @param value - string - standard, don't need to act on it
function buildObject( key, value ) {

  var allKeys = key.split("-");
  var container, i, n;
  for (container = someObj, i = 0, n = allKeys.length; i < n - 1; ++i) {
    var keyPart = allKeys[i];
    container = Object.hasOwnProperty.call(container, keyPart)
        ? container[keyPart] : (container[keyPart] = {});
  }
  container[allKeys[n - 1]] = value;
}

answered May 4, 2011 at 0:13

Mike Samuel

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4 Comments

Joshua Cody Over a year ago

Wow, magical. Any reason you used call instead of doing container.hasOwnProperty(keyPart)? Also, if it's possible to explain simply without too much effort, why alias someObj instead of extending it directly?

Mike Samuel Over a year ago

@Joshua Cody, I use Object.hasOwnProperty.call so that buildObject("hasOwnProperty", 42) doesn't break all subsequent calls to buildObject.

Mike Samuel Over a year ago

@Joshua Cody, to answer the second part of your question, container always points to the object that should contain allKeys[i] so it only aliases someObj the first time through the loop. If I didn't use a separate variable to track it, I would lose the root by the time I got to the end of the loop.

Joshua Cody Over a year ago

Awesome, Mike. I caught on to what you were doing finally once I went through the code a few times. So that's the actually magic of the solution, making sure you update the variable to point at the current location. Again, thanks so much. Your solution is helpful and elegant.

James Khoury · Accepted Answer · 2011-05-04 01:04:52Z

I came up with http://jsfiddle.net/XAn4p/ before i saw Mike's answer. I'm posting just for another way.

var newObj = function ()
    {};
newObj.prototype = 
    {
        addToObject: function (keys, value)
        {
            var keySplit = keys.split("-",2);
            if (keySplit.length  > 1)
            {
                if(this[keySplit[0]] == null)
                {
                   this[keySplit[0]] = new newObj();
                }
                var newKeys = keys.substr(keySplit[0].length +1);
                this[keySplit[0]].addToObject(newKeys, value);
            }
            else
            {
               this[keySplit[0]] = value 
            }
        }
    };

var obj = new newObj();

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