如何将特定的键值元素保留在数组中并删除其余元素？
$array_one
Array (
[2018] => Array ()
[2019] => Array
(
[5] => 2966
)
[2020] => Array
(
[0] => 2930
[1] => 2919
)
)
第二个数组是：
$成本
Array
(
[2018] => Array
(
[2789] => Array
(
[jh] => 0
[presta] => 0
[log] => 0
)
[2874] => Array
(
[jh] => 0.5
[presta] => 1
[log] => 0
)
[3786] => Array
(
[jh] => 7
[presta] => 0
[log] => 0
)
[315] => Array
(
[jh] => 0
[presta] => 0
[log] => 0
)
[325] => Array
(
[jh] => 0
[presta] => 0
[log] => 0
)
[3793] => Array
(
[jh] => 0.5
[presta] => 1.2
[log] => 0
)
[3796] => Array
(
[jh] => 22
[presta] => 27.4
[log] => 0
)
[3798] => Array
(
[jh] => 0
[presta] => 0
[log] => 0
)
[3800] => Array
(
[jh] => 17
[presta] => 0
[log] => 0
)
[3832] => Array
(
[jh] => 2
[presta] => 9
[log] => 0
)
)
我只想将那些记录保留在第一个数组中的第二个数组中，并且我想取消设置第二个数组中的剩余记录
如何仅保留与第一个数组中的键值匹配的第二个数组中的那些记录？
慕的地108432回答
-
跃然一笑
这是此答案中代码的循环版本：$output = array();foreach ($array_one as $key => $keys) { if (count($keys)) $output[$key] = array_intersect_key($costs[$key], array_flip($keys));}print_r($output);输出：Array( [2019] => Array ( [2966] => Array ( [jh] => 0 [presta] => 0 [log] => 0 ) ) [2020] => Array ( [2930] => Array ( [jh] => 0 [presta] => 0 [log] => 0 ) [2919] => Array ( [jh] => 0 [presta] => 0 [log] => 0 ) ) )3v4l.org 上的演示 -
摇曳的蔷薇
您可以使用array_intersect_key，foreach($costs as $k => &$val) // & to save value to value itself{ if(!empty($array_one[$k])){ $val = array_intersect_key($val, $array_one[$k]); }}这将减少迭代次数，$result=[];foreach ($array_one as $k => $val) { if (!empty($val)) $result[$k] = array_intersect_key($costs[$k], array_flip($val));}print_r($result);最终，它只会保留那些匹配的索引数据 array_one
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