Question

I have jQuery included so if it helps to use it, it’s available.

First the simple question:

Is there a way to check if several vars are all equal to each other?

I can use the transitive relation logic and do

if ((a == b) && (b == c)) && (c == d)) ... {

to avoid checking every variable against EACH other, but I think there should be a fancier way to do this.

If you can answer this first part only, it would be much appreciated.

Now, the tricky part…
I have a variable amount of variables (between 1 and 5)
I know that their value can be any of 200 possible values from a DDBB.

What would be the best way to know how many instances of each value I have within those variables?

For example…

If I have…

var1 = VALUE_A;
var2 = VALUE_A;
var3 = VALUE_B;
var4 = VALUE_Z;
var5 = VALUE_Z;

… i want to get something like:

result["VALUE_A"] => 2
result["VALUE_B"] => 1
result["VALUE_Z"] => 2

///////////////////////////
OR if i have…

var1 = VALUE_A;
var2 = VALUE_C;
var3 = VALUE_B;

… get:

result["VALUE_A"] => 1
result["VALUE_C"] => 1
result["VALUE_B"] => 1

///////////////////////////
OR if i have…

var1 = VALUE_A;
var2 = VALUE_A;
var3 = VALUE_A;
var4 = VALUE_C;
var5 = VALUE_C;

… get:

result["VALUE_A"] => 3
result["VALUE_C"] => 2

///////////////////////////
OR if i have…

var1 = VALUE_A;
var2 = VALUE_A;
var3 = VALUE_A;

… get:

result["VALUE_A"] => 3

Hope I was clear. Examples were the only way I could think of explaining clearly.
If this is too complex for Javascript or processing so many possible values up to times 5 can make the browser slow I can do it in PHP and get the result via AJAX but I’d rather not.

Answer 1

Would something like this do?

function countOccurrences(arr) {
    var result = {};
    $.each(arr, function(index, value) {
        if (!result[value])
            result[value] = 1;
        else
            result[value]++;
    });
    return result;
}

This function accepts an array of the values, and returns an object whose keys are the elements and values are the number of occurrences of each.