我很难想出这个问题的解决方案(可能是因为今天是星期一)。我有多个用户成员数组(电子邮件地址)，我想把它们分成唯一的组。
数据的例子:$members = array(
'Group One' => array(
'User A',
'User B',
'User C',
'User D'
),
'Group Two' => array(
'User A',
'User B',
'User D'
),
'Group Three' => array(
'User A',
'User E'
)
);
现在我想把它们分成不同的组，得到这样的结果:
Array (
0 => Array (
'Groups' => Array (
0 => 'Group One'
),
'Members' => Array (
0 => 'User C'
)
),
1 => Array (
'Groups' => Array (
0 => 'Group One',
1 => 'Group Two'
),
'Members' => Array (
0 => 'User B',
1 => 'User D'
)
),
2 => Array (
'Groups' => Array (
0 => 'Group One',
1 => 'Group Two',
2 => 'Group Three'
),
'Members' => Array (
0 => 'User A'
)
),
3 => Array (
'Groups' => Array (
0 => 'Group Three'
),
'Members' => Array (
0 => 'User E'
)
)
)
看起来这是你想要的。我们将所有组的唯一组合放入$pool数组中。在getValues函数中，我们从$a中的池(与数组相交)中获得存在于每个组中的值，并且我们在$b中获得来自其他组(与数组合并)的所有值。然后使用array_diff.
返回池中每个组中存在的用户，而不是其他组中的用户。<?php
$members = array(
'Group One' => array(
'User A',
'User B',
'User C',
'User D'
),
'Group Two' => array(
'User A',
'User B',
'User D'
),
'Group Three' => array(
'User A',
'User E'
)
);
$keys = array_keys($members);
$len = count($keys);
function getValues(&$result = array(), $members, $pool) {
$a = null;
$b = array();
foreach ($members as $group => $values) {
if (in_array($group, $pool)) {
$a = (null === $a) ? $values : array_intersect($a, $values);
} else {
$b = array_merge($b, $values);
}
}
if ($ret = array_diff($a, $b)) {
$result[] = array(
'Groups' => $pool,
'Members' => array_values($ret),
);
}
}
for ($i = 0; $i < $len; ++$i) {
$pool = array($keys[$i]);
for ($j = $i; $j < $len; ++$j) {
if ($j > $i) {
$pool[] = $keys[$j];
}
getValues($result, $members, $pool);
}
}
print_r($result);
输出如下:
Array
(
[0] => Array
(
[Groups] => Array
(
[0] => Group One
)
[Members] => Array
(
[0] => User C
)
)
[1] => Array
(
[Groups] => Array
(
[0] => Group One
[1] => Group Two
)
[Members] => Array
(
[0] => User B
[1] => User D
)
)
[2] => Array
(
[Groups] => Array
(
[0] => Group One
[1] => Group Two
[2] => Group Three
)
[Members] => Array
(
[0] => User A
)
)
[3] => Array
(
[Groups] => Array
(
[0] => Group Three
)
[Members] => Array
(
[0] => User E
)
)
)
这可能不是最好的解决方案，但应该可以工作。首先，创建一个助手数组，按用户(而不是按组)对数据进行分组，以找出每个用户的组组合。然后，遍历结果数组并构建结果。
$members = array(
'Group One' => array(
'User A',
'User B',
'User C',
'User D'
),
'Group Two' => array(
'User A',
'User B',
'User D'
),
'Group Three' => array(
'User A',
'User E'
)
);
$uniqueGroups = array();
$groupedByUser = array();
foreach($members as $group => $users) {
foreach($users as $user) {
if(!is_array($groupedByUser[$user])) {
$groupedByUser[$user] = array();
}
$groupedByUser[$user][] = $group;
}
}
foreach($groupedByUser as $user => $groups) {
sort($groups);
$found = FALSE;
foreach($uniqueGroups as $idx=>&$infos) {
$currentGroups = $infos['Groups'];
$diff = array_diff($currentGroups, $groups);
if(empty($diff)) {
$infos['Users'][] = $user;
$found = TRUE;
}
}
if(!$found) {
$uniqueGroups[] = array(
'Groups' => $groups,
'Users' => array($user)
);
}
}
print_r($uniqueGroups);