对这个数组进行分组和求和

Question

I have an array that's output is this:

Array ( [winners] => 
Array ( [0] => Gold Member 
        [1] => CROTCH SNIFFER 
        [2] => TEAM #1 )         
        [prizeTotal] => 20 )

Array ( [winners] => 
Array ( [0] => TEAM #1  
        [1] => CROTCH SNIFFER )
        [prizeTotal] => 60 ) 

Array ( [winners] => 
Array ( [0] => Gold Member 
        [1] => TEAM #1 ) 
        [prizeTotal] => 30 )

Array ( [winners] => 
Array ( [0] => TEAM #1 
        [1] => TEAM #2
        [2] => SCREW-NUT-BOLT )
        [prizeTotal] => 90 ) 

1. Please forgive the names...it's not my DB.
2. I can not change the way the array is show here.

With that being said how can I group and sum? 1. For each winners array there is a prizeTotal below the team names. That prize total should be the value of each teamName above it.

Example

Array ( [winners] => 
Array ( [0] => Gold Member 
        [1] => CROTCH SNIFFER 
        [2] => TEAM #1 )         
        [prizeTotal] => 20 )
Gold Member should have 20
CROTCH SNIFFER should have 20
TEAM #1 should have 20 AND

Array ( [winners] => 
Array ( [0] => TEAM #1  
        [1] => CROTCH SNIFFER )
        [prizeTotal] => 60 ) 
TEAM #1 should have 60
CROTCH SNIFFER sould have 60....etc....

1. Then I want to group by team name and sum so that I can display...

   CROTCH SNIFFER = 80 Gold Member = 50 TEAM #1 = 200 Team #2 = 90.

Thanks in advance...

Answer 1

Assuming you can collect all your arrays into one ie :

$arrays = array($array1,$array2,....,$arrayn);

then

$grouping = array();

foreach($arrays AS $array)
{
   foreach($array['winners'] AS $k=>$v)
   {
      //check if the array key is a number, will contain a team, and if that
      //them is not alreay listed
      if(is_numeric($k) && !array_key_exists($v,$grouping))
         $grouping[$v] = 0;
      //sum the prize to the team's sum
      $grouping[$v] += intval($array['winners']['prizeTotal']);
   }
}
//the following is just for debugging
print_r($grouping);

this should produce something like:

$grouping[TEAM #1] = 200
$grouping[TEAM #2] = 90
$grouping[CROTCH SNIFFER] = 200
...

Answer 2

I have made a function that imitates mysql SUM() and GROUP BY . I hope it will fit your needs:

           $in_a = array(
                array("a" => 0,"b"=>0,"s"=> 1),
                array("a" => 0,"b"=>0,"s"=> 2),
                array("a" => 1,"b"=>1,"s"=> 1),
                array("a" => 0,"b"=>1,"s"=> 1),
                array("a" => 0,"b"=>1,"s"=> 1),
                array("a" => 1,"b"=>0,"s"=> 1),         
                array("a" => 0,"b"=>1,"s"=> 1),         
                array("a" => 1,"b"=>1,"s"=> 1),         
                array("a" => 1,"b"=>0,"s"=> 1),         
            );//input array exaple
            $group_by_a = array("a","b");//input array will be grouped by these
            $sum_a = array("s"); //'s' values of input will be summed
            $out_a = array(); //this is the output array

            foreach($in_a as $in_i => $in)
            {
                $add = false;
                foreach($out_a as $out_i => $out)
                {
                    $add = true;
                    foreach($group_by_a as $group_by)
                        if($in[$group_by] != $out[$group_by])
                        {
                            $add = false; 
                            break;
                        }                   
                    if($add)
                    {
                        foreach($sum_a as $sum)
                            $out_a[$out_i][$sum] += $in[$sum];  
                        break;
                    }
                }
                if(!$add)
                {
                    foreach($group_by_a as $group_by)
                        $out_a[$in_i][$group_by] = $in[$group_by];
                    foreach($sum_a as $sum)
                        $out_a[$in_i][$sum] = $in[$sum];                    
                }                   
            }

the result:

Array
(
    [0] => Array
        (
            [a] => 0
            [b] => 0
            [s] => 3
        )

    [2] => Array
        (
            [a] => 1
            [b] => 1
            [s] => 2
        )

    [3] => Array
        (
            [a] => 0
            [b] => 1
            [s] => 3
        )

    [5] => Array
        (
            [a] => 1
            [b] => 0
            [s] => 2
        )

)

Answer 3

You need to make a new array to hold your results, then for each of your existing arrays, check if you're already processed one for that team

If you have, then simply add the prizeTotal to the prizeTotal stored in your new array's entry for that team.

If not, simply add the team's entry to your new array.