Two things. One, create arrays like this:

var tab = [];

instead of:
var tab = new Array()

Secondly, what you’ve created there looks like an associative array, but is actually an array with no members and the properties of f1 and f2. You can accomplish the same thing by creating an object and setting its properties.

var tab = {};
tab.f1 = “foo”;
tab.f2 = “bar”;

In other words the following two statements are identical:

tab.f1 = “foo”;
tab['f1'] = “foo”;

The main difference in usage being that you can use a variable property name using the second notation.

var baz = ‘f1′;
var tab[bav] = ‘foo’; //sets tab.f1 to “foo”

Now, to answer your question, you can iterate through them by doing:

for (each in tab) {
console.log(each);
console.log(tab[each]);
}

There are lots of quid pro quo’s in doing that, just google around.

function $A(array) {
return $.protify(array)
}

$A(myArray).findAll(…) looks a little better than $.protify(myArray).findAll(…)

var arr = [1,2,3,4,5,1,2,3,4,4,4,3,2,1];
var protArray = $.protify(arr).uniq();

Yields:
array.include is not a function
if (0 === index || (sorted ? arr…t() != value : !array.include(value))) {

If I modify ln 297 to protify (array) permanently, it works fine.

Great code. Thanks a million.