Count page's with color label

Question

Dear people,
Is it possible that InDesign can count the pages with a color label? (CS5)
So a window appears with the message.
10 yellow
20 red
etc. etc.
I' was searching arround the forum and cannot find anything with colorlabels.
Thanks for anyone who want to help!
Greetings from Holland

Answer 1

Good one, Marijan! I was thinking of an associative array as well, but I could not have written it as concise as you did.
For the inevitably flabbergasted lurkers:
a. You can use arrays like this: element[0], element[1], .. element[n], which is the regular way of addressing individual elements, but you can also address them associative: element['value'], element['name'], element['whatever']. You can use whatever you like as the index. This way, you don't have to remember '0 is for yellow, 1 is for red, 2 is for blue' -- you simply use element['yellow'], element['red'] and element['blue']. The associativity has nothing to do with the actual contents of each array element.
b. So Marijan uses 'pageColor' to store the counter for each page color -- if you encounter a page with a color label 'red', he adds 1 to pC['red'].
c. Problem is, you cannot add 1 to something that does not exist. If you encounter a page color 'purple' for the first time, you cannot say
pC['purple'] = pC['purple']+1;
(or a shorthand equivalent of the same) because it doesn't exist yet!
d. A non-existing array element is equal to null -- "nothing". (This is not the same as '0' (zero).) Now while you can check for
if (pC['purple'] == null) ..
you can also use the logical operator ! (not), which tests for equal-to-zero (and in this case, null is treated as if it is 0 -- that actually makes sense because it's certainly not not zero). The line
!(object)
will result in true if object equals null, false otherwise -- !x tests for zero, so if x is zero, the result is true. Capice?
e. By way of encore Marijan packs the entire test-if-exist/create-new/add-one into a single line. The notation
a ? b : c;
works like this: if (a) is true then do (b) else do (c). So if 'new color' is null -- it doesn't exist -- then add a new associated element (and set its counter value to '1', by the way), else add '1' to the value it already had.