Lazy, lazy, lazy: concatenate it in an ordinary array, sort that and split the values into a new associative array. Probably not recommended for the phone book but like a say - works for me.

Also removes duplicates

function sortAssoc(inputArray) {
var tempArray = new Array();
var outputArray = new Array();

i=0;

for (keyVar in inputArray) {
    tempArray[i] = inputArray[keyVar] + '//--string_divider--//' + keyVar;
    i++;
}

tempArray = tempArray.sort();

for (i=0; i < tempArray.length; i++) {
    if (tempArray[i]!=tempArray[i-1]) {
        temp = tempArray[i].split('//--string_divider--//');
        tempIndex = temp[1];
        outputArray[tempIndex] = temp[0];
    }
}

return outputArray;

}

An intriguing problem, although I’m not sure I understand what you’re trying to do so I hope my answer makes some sense.

I don’t think JavaScript has this functionality built-in. It may be possible to build a user-defined function that would do this but the simplest solution I can think of is to keep an ordinary array of the hash table keys. Something like this:

var keyIndex = new Array();
var hashTable = new Object();

hashTable[”z”] = “the last item”;
keyIndex.push(”z”);

hashTable[”x”] = “the first item”;
keyIndex.push(”x”);

hashTable[”y”] = “the middle item”;
keyIndex.push(”y”);

You can then use the Array() object’s built-in sort() function to order the keys and loop through the keyIndex array to give you your keys in order:

keyIndex.sort();

var i;
for (i in keyIndex)
{
document.write(hashTable[ keyIndex[i] ] + “<br />”);
}

This should produce the output:

the first item
the middle item
the last item

There’s a bit of overhead in managing both the keyIndex and hashTable objects so I don’t think it will be particularly fast. And I don’t think it’s a very elegant solution but at least it’s an option for you. I’m sorry if it’s not really what you’re looking for.

Please note: I haven’t tested this example code, so no guarantees that it’ll work!

Cheers,

Trevor.

What if you WANT your elements to have a logical sequence (say, alphabetical). For example,

var map = new Object();

var foo = [x, y, z];
var bar = [v, w, z];
var baz = [a, b, c];

I want to be able to add foo to map, and then bar, and have bar be “before” foo (because it comes first alphabetically). If I later add baz, I want it to be “between” bar and foo for the same reason.

I get why associative arrays don’t have any logical sequence, and so adding ‘in sequence’ doesn’t really make sense. What I’m describing (and wanting!) is more like the Java TreeMap-like objects, which maintains the keys in order and guarantees quick access (I need both).

What would solve this for me is a way to sort() the keys of an associative array/object (maintaining the associated element relationships of course).

I’ve searched for both these solutions but haven’t found one yet. I thought I’d give you a try and see if you had any ideas.

Thanks in advance!

e.

Hi Eugene,

As you suggest, a multidimensional hash can be made by making a series of objects inside the first hash.

varName[1] = new Object();
varName[2] = new Object();
... and so on

That’s a perfectly acceptable way to do it.

However, if you find that’s a bit of overkill you can also fake it just by concatenating the key values:

var varName = new Object()

var i = 1 + "z" + 33;

varName[i] = 1;
varName[1 + "y" + 34] = 2;
varName[2 + "aa" + 'a'] = 3;
varName[2 + "aa" + 'b'] = 4;
varName[2 + "aa" + 'c'] = 5;

i = "2aaa";
alert(varName["1z33"]);
alert(varName[i]);
alert(varName[2 + "aa" + 'c']);

If you prefer your multiple dimensions separated, say by a comma, just include it in the concatenation:

varName[1 + "," + "z" + "," + 33] = 1;

alert(varName["1,z,33"]);

I’d recommend this both for readability and especially if you’re using numeric keys (remember that 1 + 2 + 3 is not the concatenated value “123″, it’s the numeric value 6).

Sure it’s a hack, but it’s legit: it doesn’t matter how you make your hash indexes as long as each one is unique.

Cheers,

Trevor.

varName[1]["z"][33] = 1;
varName[1]["y"][34] = 2;
varName[2]["aa"]['a'] = 3;
varName[2]["aa"]['b'] = 4;
varName[2]["aa"]['c'] = 5;

without varName[1] = new Object();