php中foreach的一个问题
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2022-05-11 15:28:57
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$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 ))