[英]How to check whether an array of array have a key value pair in its values in PHP or Laravel
I have the following array. 我有以下数组。 I want to know whether any of the values of the array contains the key-value pair
"RoleCode" => "Admin"
. 我想知道数组的任何值是否包含键值对
"RoleCode" => "Admin"
。
[
0 => [
"RoleCode" => "Admin"
"RoleName" => "Administrator"
]
1 => [
"RoleCode" => "PM"
"RoleName" => "ProjectManager"
]
2 => [
"RoleCode" => "ScheduleUser"
"RoleName" => "Schedule User"
]
]
I can write a long code to find it out like the following: 我可以编写一个长代码来查找它,如下所示:
$isAdmin = false;
foreach ($user['Roles'] as $role) {
if ($role['RoleCode'] == 'Admin') {
$isAdmin = true;
}
}
Is there any way to do this in a better way? 有什么办法可以更好地做到这一点?
You could use array_column()
and in_array()
: 您可以使用
array_column()
和in_array()
:
$isAdmin = in_array('Admin', array_column($user['Roles'], 'RoleCode')) ;
array_column()
will return an array with all values from 'RoleCode'
key array_column()
将返回具有'RoleCode'
键的所有值的数组 in_array()
will check if Admin
is inside in_array()
将检查Admin
是否在内部 It depends what is better way. 这取决于哪种更好的方法。
Current solution with adding break
when item found: 找到项目时添加
break
当前解决方案:
$isAdmin = false;
foreach ($user['Roles'] as $role) {
if ($role['RoleCode'] == 'Admin') {
$isAdmin = true;
break;
}
}
will be O(n)
in worst case. 在最坏的情况下将为
O(n)
。
Other solutions, like one in another answer 其他解决方案,如另一个答案
$isAdmin = in_array('Admin', array_column($user['Roles'], 'RoleCode'));
This will be O(n) + O(1)
in best case and O(n) + O(n)
in worst. 这将是
O(n) + O(1)
在最好的情况下和O(n) + O(n)
在最差。 More than initial foreach
. 不仅仅是最初的
foreach
。
Another one is filtering: 另一个是过滤:
$isAdmin = !empty(array_filter(
$user['Roles'],
function ($v) { return $v['RoleCode'] == 'Admin'; }
));
It is always O(n)
总是
O(n)
So, from the point of readability and performance, initial code is the winner. 因此,从可读性和性能的角度来看,初始代码是赢家。
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