php中foreach的一個有關問題

php中foreach的一個問題

$list = array(
            array('id' => 1, 'count' => 1),
            array('id' => 1, 'count' => 2),
            array('id' => 2, 'count' => 2),
            array('id' => 3, 'count' => 3),
            array('id' => 3, 'count' => 4)
        );
        $sort = array();
        foreach($list as $v)
        {
            $sort[] = $v['id'];
        }
        $sort = array_unique($sort);
        $new = array();
        foreach($sort as $v1)
        {
            $c = 0;
            foreach($list as $v2)
            {
                if($v1 == $v2['id'])
                {
                    $c += $v2['count'];
                }
            }
            $new[] = array('id' => $v1, 'count' => $c);
        }

這是現在的做法，求簡單更合理的做法！！
需求：現有的數組：$list,需要得到的結果：把數組內重複的id對應的欄位count合并。如上面的結果數組$new

------解決思路----------------------

$list = array(
            array('id' => 1, 'count' => 1),
            array('id' => 1, 'count' => 2),
            array('id' => 2, 'count' => 2),
            array('id' => 3, 'count' => 3),
            array('id' => 3, 'count' => 4)
        );

// 執行計算操作
$tmp = array();

foreach($list as $val){
    if(isset($tmp[$val['id']])){
        $tmp[$val['id']] += $val['count'];
    }else{
        $tmp[$val['id']] = $val['count'];
    }
}

ksort($tmp); // 按key排序

$new = array();
foreach($tmp as $k=>$v){
    array_push($new,array('id'=>$k,'count'=>$v));
}

print_r($new);

------解決思路----------------------

$list = array(
  array('id' => 1, 'count' => 1),
  array('id' => 1, 'count' => 2),
  array('id' => 2, 'count' => 2),
  array('id' => 3, 'count' => 3),
  array('id' => 3, 'count' => 4)
);
$new = array();
foreach($list as $r) {
  if(! isset($new[$r['id']])) $new[$r['id']] = $r;
  else $new[$r['id']]['count'] += $r['count'];
}
$new = array_values($new);
print_r($new);

Array
(
    [0] => Array
        (
            [id] => 1
            [count] => 3
        )

    [1] => Array
        (
            [id] => 2
            [count] => 2
        )

    [2] => Array
        (
            [id] => 3
            [count] => 7
        )

)

