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);
