Php 仅基于新键合并两个多维数组
有没有合并这些阵列的方法:Php 仅基于新键合并两个多维数组,php,array-merge,Php,Array Merge,有没有合并这些阵列的方法: // saved settings by the user $arr1 = [ 'options' => [ 'first' => 1, 'second' => 2, 'third' => 3, ] ]; // new option schema (new keys added) $arr2 = [ 'options' => [ 'first' =&
// saved settings by the user
$arr1 = [
'options' => [
'first' => 1,
'second' => 2,
'third' => 3,
]
];
// new option schema (new keys added)
$arr2 = [
'options' => [
'first' => 1,
'second' => 212,
'fourth' => 4
]
];
并得到如下输出:
$arr3 = [
'options' => [
'first' => 1, // nothing do to, "first" key already exists
'second' => 2, // nothing to do, "second" key exists (user already saved "second" with value 2, so we omit value 212)
// "third" option got removed from new schema, no longer needed in the app, so may be removed from User settings as well
'fourth' => 4 // this key is new, so let's add it to the final result
]
];
基本上,我尝试了array\u merge
或array\u merge\u recursive
但是它们合并了所有键,而不是只合并新键,因此它覆盖了用户设置
当然,源数组要复杂得多,里面有很多多维数组
有什么方法可以使它变得简单,或者库可以处理它吗?可以使用递归函数来完成。新结构(
$arr2
在本例中)定义了结果中存在的键。如果旧结构在新结构的相应键处具有值,则将使用该值。否则,将使用新结构中的值。因为您只查看新结构中存在的键,所以不会包括旧结构中不再存在的任何键
function update($newKeys, $oldData) {
$result = [];
// iterate only new structure so obsolete keys won't be included
foreach ($newKeys as $key => $value) {
// use the old value for the new key if it exists
if (isset($oldData[$key])) {
if (is_array($oldData[$key]) && is_array($value)) {
// if the old and new values for the key are both arrays, recurse
$result[$key] = merge($value, $oldData[$key]);
} else {
// otherwise, just use the old value
$result[$key] = $oldData[$key];
}
// use the new value if the key doesn't exist in the old values
} else {
$result[$key] = $value;
}
}
return $result;
}
$result = update($arr2, $arr1);
我只会在完全关联的结构上使用它。如果任何内部数组都是键不重要的基本索引数组,则必须向函数中添加一些内容,以避免将它们弄乱。您希望在
$arr3
中使用的不是合并的概念,您需要通过循环或类似的构造手动执行该操作Hanks Abdo,我想可能已经有现成的解决方案了,如果没有,我会自己做的