当给定父数组（关联数组）的键名时，是否有一个我错过的PHP函数会更改父数组的键，或者是否至少有一个foreach循环的替代方案，我目前正在使用它来更改键。
阵列示例
$arr = array(
array(
'id' => 1,
'name' => 'one',
),
array(
'id' => 2,
'name' => 'two',
),
array(
'id' => 3,
'name' => 'three',
)
);
我希望它能像这样工作…
$arr_name = array_change_key($arr,'name');
print_r($arr_name);
$arr_name => array(
'one', => array(
'id' => 1,
'name' => 'one',
),
'two' => array(
'id' => 2,
'name' => 'two',
),
'three' => array(
'id' => 3,
'name' => 'three',
)
);
//$arr is unchanged
这只是一个额外的（不确定是否可能）
array_change_key($arr,'name');
print_r($arr);
//$arr has changed because it doesn't have a variable to set
$arr => array(
'one', => array(
'id' => 1,
'name' => 'one',
),
'two' => array(
'id' => 2,
'name' => 'two',
),
'three' => array(
'id' => 3,
'name' => 'three',
)
);
print_r($arr[0]); //undefined index
如果我正确理解了这个问题，比如：
$arr = array_combine(
array_column($arr, 'name'),
$arr
);
将使用每条记录中的name值作为父密钥，并给出
array(3) {
["one"]=>
array(2) {
["id"]=>
int(1)
["name"]=>
string(3) "one"
}
["two"]=>
array(2) {
["id"]=>
int(2)
["name"]=>
string(3) "two"
}
["three"]=>
array(2) {
["id"]=>
int(3)
["name"]=>
string(5) "three"
}
}
您必须告诉函数是否"通过引用传递"，它无法知道您是否试图将返回的结果设置为变量；
function array_change_key(array &$array, $key, $pass_by_reference = false){
if($pass_by_reference){
// check is_scalar($key)
if(!is_scalar($key)) return FALSE;
// we already know isset($array), is_array($array) and isset(key) are true because $pass_by_reference is true;
$array = markBakersAnswer($array,$key);
return TRUE;
// pass-by-reference functions usually return true or false
}
return markBakersAnswer($array,$key);
}
MarkBakersAnswer+1
$new_array = array_change_key($arr, 'name'); // $arr unchanged and $new_array == array with new keys
$new_array = array_change_key($arr, 'name', false); // $arr unchanged and $new_array == array with new keys
$new_array = array_change_key($arr, 'name', true); // $arr changed (new keys), $new_array = TRUE;
$new_array = array_change_key($arr, array(), true); // $arr changed (new keys), $new_array = FALSE;