[RESOLVED] Two Arrays, One Common Entry

BryanJ

I have two arrays, one of which is skillInTraining which consists of these entries:

Array

(
    [currentTQTime] => 2009-01-25 01:06:20
    [trainingEndTime] => 2009-01-26 09:22:01
    [trainingStartTime] => 2009-01-23 17:29:13
    [trainingTypeID] => 3393
    [trainingStartSP] => 8000
    [trainingDestinationSP] => 45255
    [trainingToLevel] => 4
    [skillInTraining] => 1
)

and the other array, which is skill tree. I would paste the skill tree Array but it is too big. You can view it at: http://pastebin.ca/1317566

The two arrays have one entry in common: in the first array it is trainingTypeID and in the second array it is TypeID. Is there a way to compare the two arrays and output the value of typeName for the part of the second array that matches the first?

BryanJ

Anyone?

NogDog

Your pastebin.ca link does not work.

BryanJ

I cant access the array right now, but the only thing they have in common is typeID and trainingTypeID. Is there a way to output the information in the part of the array where the two numbers match?

BryanJ

Anyone have any idea from this?

The other array is http://pastebin.com/m72385c78in addition to the http://pastebin.ca/1317566 one which is now working. The pastebin.com link is only an except that shows half of the array, but it should be enough to get a general look on how the array is setup

NogDog

/**
 * @return mixed value if found, else boolean false
 * @param array $arr
 * @param string $key1 Array key for which to find specified value
 * @param string $value value to be searched for for key1
 * @param string $key2 key whose value is to be returned by this function
 **/
function findit($arr, $key1, $value, $key2)
{
   if(isset($arr[$key1]) and $arr[$key1] == $value and isset($arr[$key2]))
   {
      return $arr[$key2];
   }
   else
   {
      foreach($arr as $val)
      {
         if(is_array($val))
         {
            $result = findit($val, $key1, $value, $key2);
            if($result !== false)
            {
               return $result;
            }
         }
      }
   }
   return false;
}

// usage:
$typeName = findit($skillTree, 'typeID', $skillInTraining['trainingTypeID'], 'typeName');

BryanJ

Oh thank god! You are truly my savior!