返回两个数组的交集
A数组
- PHP code
array( [0] => array( ['action_id'] => 3 ) [1] => array( ['action_id'] => 2 ) [2] => array( ['action_id'] => 1 ) [3] => array( ['action_id'] => 7 ) [4] => array( ['action_id'] => 11 ))
B数组：
- PHP code
array( [0] => array( ['action_id'] => 3 ['type'] => 0 ['order_num'] => 67 ) [1] => array( ['action_id'] => 2 ['type'] => 0 ['order_num'] => 66 ) [2] => array( ['action_id'] => 1 ['type'] => 0 ['order_num'] => 65 ) [3] => array( ['action_id'] => 7 ['type'] => 0 ['order_num'] => 64 ) [8] => array( ['action_id'] => 14 ['type'] => 0 ['order_num'] => 40 ) [13] => array( ['action_id'] => 11 ['type'] => 0 ['order_num'] => 30 ))
使用array_intersect_key返回的为什么是
- PHP code
array( [0] => array( ['action_id'] => 3 ['type'] => 0 ['order_num'] => 67 ) [1] => array( ['action_id'] => 2 ['type'] => 0 ['order_num'] => 66 ) [2] => array( ['action_id'] => 1 ['type'] => 0 ['order_num'] => 65 ) [3] => array( ['action_id'] => 7 ['type'] => 0 ['order_num'] => 64 ) )
action_id为11的为什么没了？？
------解决方案--------------------
- PHP code
$a = array( 0 => array('action_id' => 3), 1 => array('action_id' => 2), 2 => array('action_id' => 1), 3 => array('action_id' => 7), 4 => array('action_id' => 11),);$b = array( 0 => array('action_id' => 3, 'type' => 0, 'order_num' => 67), 1 => array('action_id' => 2, 'type' => 0, 'order_num' => 66), 2 => array('action_id' => 1, 'type' => 0, 'order_num' => 65), 3 => array('action_id' => 7, 'type' => 0, 'order_num' => 64), 8 => array('action_id' => 14, 'type' => 0, 'order_num' => 40), 13 => array('action_id' => 11, 'type' => 0, 'order_num' => 30),);foreach($a as $v) $dict[] = $v['action_id'];foreach($b as $k=>$v) if(in_array($v['action_id'], $dict)) $c[$k] = $v;print_r($c);
