我有一个包含事件和日期的二维数组:
$events = array(
array('event' => 'event1', 'date' => '2016-05-05'),
array('event' => 'event2', 'date' => '2016-05-08'),
array('event' => 'event3', 'date' => '2016-05-08'),
array('event' => 'event4', 'date' => '2016-05-10'),
array('event' => 'event5', 'date' => '2016-05-10'),
array('event' => 'event6', 'date' => '2016-05-11'),
array('event' => 'event7', 'date' => '2016-05-11'),
array('event' => 'event8', 'date' => '2016-05-13')
};
让我们说今天是2016-05-10
,我想创建3个新数组:
$ recent - 所有事件都发生在上一个可用日期 $ today - 今天所有活动
$ coming - 所有活动在下一个可用日发生
$today
很简单:
$today = array();
for($i = 0; $i < count($events); ++$i) {
if($events[$i]['date'] == date("Y-m-d") {
array_push($today, $events[$i]['event']);
}
}
因此,我需要$recent
包含event2
和event3
,以及$upcoming
来包含event6
和event7
。
问题是如何查找recent
和upcoming
。
*澄清:
我不希望$recent
中的所有事件都发生在今天之前,但事件发生在前一天。因此,在这种情况下,只有2016-05-08
答案 0 :(得分:3)
// Take all dates from source array
$dates = array_unique(array_map(function ($i) { return strtotime($i); } , array_column($events, 'date')));
sort($dates);
$today = strtotime('midnight');
// find previouse date. It will be 1970-1-1 if not present in array
$prev = @max(array_filter($dates, function($i) use($today) { return $i < $today; }));
// find туче date. It will be 1970-1-1 if not present in array
$next = @min(array_filter($dates, function($i) use($today) { return $i > $today; }));
$prev = date('Y-m-d', $prev);
$next = date('Y-m-d', $next);
// fill arrays
$recent = array();
$upcoming = array();
$today = array();
for($i = 0; $i < count($events); ++$i) {
if($events[$i]['date'] == date('Y-m-d')) {
array_push($today, $events[$i]['event']);
}
if($events[$i]['date'] == $prev) {
array_push($recent, $events[$i]['event']);
}
if($events[$i]['date'] == $next) {
array_push($upcoming , $events[$i]['event']);
}
}
答案 1 :(得分:1)
尝试创建三(3)个空白数组,检查日期是否大于或小于今天,并根据条件推入数组。
$today = array();
$upcoming = array();
$recent = array();
$thisDay = date("Y-m-d");
$count = count($events);
$max = max(array_column($events, 'date'));
$min = min(array_column($events, 'date'));
for($i = 0; $i < $count; $i++){
if($events[$i]['date'] > $thisDay){
$max = ($max > $events[$i]['date']) ? $events[$i]['date'] : $max;
array_push($upcoming, $events[$i]['event']);
array_push($upcoming_dates, $events[$i]['date']);
}
elseif($events[$i]['date'] < $thisDay){
$min = ($min < $events[$i]['date']) ? $events[$i]['date'] : $min;
array_push($recent, $events[$i]['event']);
array_push($recent_dates, $events[$i]['date']);
}
else
array_push($today, $events[$i]['event']);
}
foreach($recent_dates as $key => $value){
if($value != $min)
unset($recent[$key]);
}
foreach($upcoming_dates as $key => $value){
if($value != $max)
unset($upcoming[$key]);
}
echo '<pre>';
print_r($today);
print_r($upcoming);
print_r($recent);
<强>结果强>
今天:
Array
(
[0] => event4
[1] => event5
)
即将来临:
Array
(
[0] => event6
[1] => event7
)
最近:
Array
(
[1] => event2
[2] => event3
)
注意:您使用的
push_array
不是PHP中的任何库函数。对于最近和即将推出的重新索引,您可以使用array_values
。
答案 2 :(得分:0)
另一个解决方案应该是
$recent = array();
$upcoming = array();
$today = array();
$all_dates = array();
foreach ($events as $event):
array_push($all_dates, $event['date']);
endforeach;
if ($key = array_search('2016-05-10', $all_dates)) {
$prev_date = $all_dates[$key - 1];
}
for ($i = 0; $i < count($all_dates); $i++) {
if ($all_dates[$i] > $all_dates[$key]) {
$next_date = $all_dates[$i + 1];
break;
}
}
for ($i = 0; $i < count($events); ++$i) {
if ($events[$i]['date'] == date('Y-m-d')) {
array_push($today, $events[$i]['event']);
}
if ($events[$i]['date'] == $prev_date) {
array_push($recent, $events[$i]['event']);
}
if ($events[$i]['date'] == $next_date) {
array_push($upcoming, $events[$i]['event']);
}
}
echo '<pre>';
print_r($upcoming);
print_r($today);
print_r($recent);
输出
Array
(
[0] => event6
[1] => event7
)
Array
(
[0] => event4
[1] => event5
)
Array
(
[0] => event2
[1] => event3
)