$array_oneArray ( [2018] => Array () [2019] => Array ( [5] => 2966 ) [2020] => Array ( [0] => 2930 [1] => 2919 ))第二个数组是：$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？
2 回答
跃然一笑
TA贡献1826条经验 获得超6个赞
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
)
)
)
摇曳的蔷薇
TA贡献1793条经验 获得超6个赞
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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