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php中foreach的一个问题

WBOY
WBOYOriginal
2016-06-20 12:58:291090browse

$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' => $c);        }这是现在的做法,求简单更合理的做法!!需求:现有的数组:$list,需要得到的结果:把数组内重复的id对应的字段count合并。如上面的结果数组$new


回复讨论(解决方案)

$list = array(            array('id' => 1, 'count' => 1),            array('id' => 1, 'count' => 2),            array('id' => 2, 'count' => 2),            array('id' => 3, 'count' => 3),            array('id' => 3, 'count' => 4)        );// ?行?算操作$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);

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

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