【发布时间】：2021-06-09 18:14:23
【问题描述】：
我正在循环一个多维数组，并留下一些值。
这是完整的 PHP 代码。
<?php
/**
* Each designer is required 3x per week,
* add their name 3 times to allow for the count.
*
* [array_pop] removes the designer from the sub-array
* u[sort($designers, "sortByTotalItems")] then moves the designers with more names remaining to the top of the list
*
* @var [type]
*/
$designers = array(
[ "Aaron Summers", "Aaron Summers", "Aaron Summers" ],
[ "Adam Smart", "Adam Smart", "Adam Smart" ],
[ "Andrew Montgomery", "Andrew Montgomery", "Andrew Montgomery" ],
[ "Dave Robertson", "Dave Robertson", "Dave Robertson" ],
[ "Holly Colby", "Holly Colby", "Holly Colby" ],
[ "Karen Fok", "Karen Fok", "Karen Fok" ],
[ "Kieran Lintott", "Kieran Lintott", "Kieran Lintott" ],
[ "Rebecca Roffey", "Rebecca Roffey", "Rebecca Roffey" ],
[ "Sonia Tam", "Sonia Tam", "Sonia Tam" ],
[ "Steffi Wallace", "Steffi Wallace", "Steffi Wallace" ],
[ "Taz Patel", "Taz Patel", "Taz Patel" ]
);
/**
* Store the original designers to reset them once empty
*
* @var [type]
*/
$designers_bak = $designers;
/**
* Generate an array of string dates between 2 dates
*
* @param string $start Start date
* @param string $end End date
* @param string $format Output format (Default: Y-m-d)
*
* @return array
*/
function getDatesFromRange( $start, $end ) {
$array = array();
$interval = new DateInterval('P1D');
$realEnd = new DateTime($end);
$realEnd->add($interval);
$period = new DatePeriod(new DateTime($start), $interval, $realEnd);
foreach($period as $date) {
$dayOfWeek = $date->format('l');
if ( $dayOfWeek != 'Saturday' && $dayOfWeek != 'Sunday' ) {
$array[] = $date->format('Y-m-d');
}
}
return $array;
}
/**
* Sort the array into sub-arrays that contain the most values
* array(
* array('name', 'name', 'name')
* array('name', 'name')
* array('name')
* )
* https://stackoverflow.com/questions/7433569/php-sort-a-multidimensional-array-by-number-of-items#answer-7433753
*/
function sortByTotalItems($a, $b) {
if ($a == $b) {
return 0;
}
return (count($a) > count($b)) ? -1 : 1;
}
/**
* Get the total amount of days in this week 1-5
*
* @var [type]
*/
$days_at_work = 3;
$total_days_in_week = 5;
$assigned_total = floor(count( $designers ) * $days_at_work / $total_days_in_week);
/**
* Each day
*
* @var [type]
*/
$working_weeks = array_chunk(getDatesFromRange('09-06-2021', '09-06-2022'), $total_days_in_week);
$given_designer = array();
foreach ( $working_weeks as $working_days ) {
/**
* Reset the designers
*
* @var [type]
*/
foreach ( $working_days as $working_day ) {
/**
* Randomise the designers
*
* @var [type]
*/
shuffle( $designers );
/**
* Sort the shuffle, so the sub-arrays count is descending
* array(
* array('name', 'name', 'name')
* array('name', 'name')
* array('name')
* )
*
* @var [$designers]
*/
$designers = array_filter(array_map('array_filter', $designers));
usort($designers, "sortByTotalItems");
/**
* Add designer to the working day
*
* @var [type]
*/
$forcount = ( count($designers) < $assigned_total ) ? count($designers) : $assigned_total;
for ($d=0; $d < $forcount; $d++) {
/**
* Add designer to the days visit
*
* @var [type]
*/
$given_designer[] = array(
"title"=> $designers[$d][0],
"start"=> $working_day
);
/**
* Remove the name from the names array -1 each loop
*
* @var [type]
*/
array_pop($designers[$d]);
/**
* remove empty arrays so we can count the remaining designers
* If there are less than 1 weeks worth of designers, reset the variable
*/
$current_designers = array_filter(array_map('array_filter', $designers));
if ( count($current_designers) < $assigned_total ) {
$designers = $designers_bak;
}
}
}
}
?>
下面是我在循环和 array_pop 之后剩下的内容。
我们称之为 array1
<?php
array ( 0 => array ( ),
1 => array ( ),
2 => array ( ),
3 => array ( ),
4 => array ( 0 => 'Aaron Summers', ),
5 => array ( 0 => 'Taz Patel', ),
6 => array ( 0 => 'Sonia Tam', ),
7 => array ( 0 => 'Adam Smart', ),
8 => array ( 0 => 'Andrew Montgomery', ),
)
?>
然后我用原始值重置数组：
我们称之为 array2
<?php
array ( 0 => array ( 0 => 'Aaron Summers', 1 => 'Aaron Summers', 2 => 'Aaron Summers', ),
1 => array ( 0 => 'Adam Smart', 1 => 'Adam Smart', 2 => 'Adam Smart', ),
2 => array ( 0 => 'Andrew Montgomery', 1 => 'Andrew Montgomery', 2 => 'Andrew Montgomery', ),
3 => array ( 0 => 'Dave Robertson', 1 => 'Dave Robertson', 2 => 'Dave Robertson', ),
4 => array ( 0 => 'Holly Colby', 1 => 'Holly Colby', ),
5 => array ( 0 => 'Karen Fok', 1 => 'Karen Fok', 2 => 'Karen Fok', ),
6 => array ( 0 => 'Kieran Lintott', 1 => 'Kieran Lintott', 2 => 'Kieran Lintott', ),
7 => array ( 0 => 'Rebecca Roffey', 1 => 'Rebecca Roffey', 2 => 'Rebecca Roffey', ),
8 => array ( 0 => 'Sonia Tam', 1 => 'Sonia Tam', 2 => 'Sonia Tam', ),
9 => array ( 0 => 'Steffi Wallace', 1 => 'Steffi Wallace', 2 => 'Steffi Wallace', ),
10 => array ( 0 => 'Taz Patel', 1 => 'Taz Patel', 2 => 'Taz Patel', ),
)
?>
我如何能够根据留在 array1 中的人重新排序 array2 中的项目？
array2 是数组和原始形式的数组的总大小，在我循环并使用 array_pop() 删除名称之前。
【问题讨论】：
-
首先，您已经在第二个数组中添加了一些数据，请注意解释那里的逻辑
-
这是数组的总大小吗？或者这在现实中可能是一个非常大的数组
-
array2 是数组的总大小，我已经用更多细节更新了我的问题
-
好吧，您将不得不分几步完成此操作。首先找到第一个数组中的所有名称并放入一个新数组中，对其进行排序，然后将每个名称的 3 次出现伪造到新数组中2
-
我想我们应该假设在 array1 中确实有 11 + 空出现，或者我们将如何设法创建丢失的 6 个名称