up vote 3 down vote favorite
1
share [g+] share [fb]

I am looking to do this a better way without the need to hardcode the integers for $justPrices[$i]:

$pricesResult = array_merge($justPrices[0], $justPrices[1], $justPrices[2], $justPrices[3]);

$justPrices is a multidimensional array, containing 4 'bands' of prices within each array. The data for $justPrices being for example:

Array ( [0] => Array ( [0] => 40.95 [1] => 39.95 [2] => 39.45 [3] => 38.95 ) [1] => Array ( [0] => 45.80 [1] => 41.80 [2] => 41.50 [3] => 41.40 ) [2] => Array ( [0] => 45.95 [1] => 42.95 [2] => 41.95 [3] => 41.45 ) [3] => Array ( [0] => 50.00 [1] => 50.00 [2] => 50.00 [3] => 50.00 ) )

The issue is that the amount of arrays within $justPrices will vary from at least 2 to 10+. So I need a way for the parameters for the array_merge() function to vary dependent on the amount of arrays within $justPrices. I was going to use this simple method to get the amount of arrays within $justPrices:

$justPricesMax = count($justPrices);

I could write a for loop, and I might still, I just wondered if there was a better method for what seems on the surface relatively simple!

Kind regards, Neil

link|improve this question
What result do you expect? – Gumbo Sep 3 '10 at 10:03
@Gumbo a single array containing all the prices. I am using that for an average price thing afterwards. – NezZ Sep 3 '10 at 10:26
feedback

1 Answer

up vote 5 down vote accepted

If you just want to flatten the array, you can use call_user_func_array to call array_merge with the elements of $justPrices as parameters:

$flat = call_user_func_array('array_merge', $justPrices);

This is equivalent to a the function call:

$flat = array_merge($justPrices[0], $justPrices[1], … , $justPrices[count($justPrices)-1]);
link|improve this answer
+1: Love the PHP kitchen sink. – shamittomar Sep 3 '10 at 10:16
That is literally it. Fantastic response. – NezZ Sep 3 '10 at 10:29
feedback

Your Answer

 
or
required, but never shown

Not the answer you're looking for? Browse other questions tagged or ask your own question.