我正在构建事件列表日历。事件具有开始日期和结束日期,日期以YYYYMMDD格式存储。活动列表按天查看,活动可以在一天内完成,也可以在多天内完成。按两个日期和规则对数组排序
活动当前存储在一个数组,start_date
和end_date
我怎样才能:
- 此举在当前日期到 阵列的顶部开始的任何事件。
- 然后按开始日期
- 对事件排序然后通过结束日期
有事件排序看到这个question,但徘徊如何建立起来的逻辑之上。
我正在构建事件列表日历。事件具有开始日期和结束日期,日期以YYYYMMDD格式存储。活动列表按天查看,活动可以在一天内完成,也可以在多天内完成。按两个日期和规则对数组排序
活动当前存储在一个数组,start_date
和end_date
我怎样才能:
有事件排序看到这个question,但徘徊如何建立起来的逻辑之上。
将当前日期开始的所有事件移动到数组的顶部,但保留其他人按原样排列!
$today = date('Ymd',strtotime('today'));
usort($events,function($a,$b) use($today){
if($a['start_date'] === $today && $b['start_date'] != $today){return -1;}
else if($b['start_date'] === $today && $a['start_date'] != $today){return 1;}
else {return strnatcmp($a['start_date'],$b['start_date']);}
});
或排序事件由起始日期
usort($events,function($a,$b){
return strnatcmp($a['start_date'],$b['start_date']);
});
或通过结束日期的事件排序
usort($events,function($a,$b){
return strnatcmp($a['end_date'],$b['end_date']);
});
编辑一些测试数据:
//example works best if today is 20170810
$events = array(
array(
'name' => 'a',
'start_date' => '20170810',
'end_date' => '20170811',
),
array(
'name' => 'b',
'start_date' => '20170810',
'end_date' => '20170810',
),
array(
'name' => 'c',
'start_date' => '20170607',
'end_date' => '20170608',
),
array(
'name' => 'd',
'start_date' => '20170607',
'end_date' => '20170607',
),
array(
'name' => 'e',
'start_date' => '20170810',
'end_date' => '20170812',
),
array(
'name' => 'f',
'start_date' => '20170807',
'end_date' => '20170817',
),
);
print_r($events);
编辑:他们都在一起:持续事件是由“被下令:事件是由 '吗?他们从今天开始', '起始日期', 'END_DATE'
$today = date('Ymd',strtotime('today'));
usort($events,function($a,$b) use($today){
//one of them starts today
if($a['start_date'] === $today && $b['start_date'] != $today){return -1;}
else if($b['start_date'] === $today && $a['start_date'] != $today){return 1;}
//both or neither start today, compare start_date
else {
if($a['start_date'] != $b['start_date']){
//start_dates differ, order by them
return strnatcmp($a['start_date'],$b['start_date']);
} else {
//start dates are the same, order by end_date
return strnatcmp($a['end_date'],$b['end_date']);
}
}
});
print_r($events);
编辑命令?他们目前正在进行中“‘起始日期’,‘END_DATE’
$today = date('Ymd',strtotime('today'));
//function compares two elements to decide their position relative to each other
usort($events,function($a,$b) use($today){
//check if event is currently ongoing
//starts or ends today or today is between start and end
$ongoing = function($event) use($today){
//this is the simplest form I could think of for the check
return (strnatcmp($event['start_date'],$today) < 1 && strnatcmp($today,$event['end_date']) < 1);
};
$startstoday = function($event) use($today){
return $event['start_date'] === $today;
};
//only one of them is ongoing
if($ongoing($a) && !$ongoing($b)){return -1;}
else if($ongoing($b) && !$ongoing($a)){return 1;}
//both or neither are ongoing, compare start_date
else {
if($a['start_date'] != $b['start_date']){
//start_dates differ, order by them
return strnatcmp($a['start_date'],$b['start_date']);
} else {
//start dates are the same, order by end_date
return strnatcmp($a['end_date'],$b['end_date']);
}
}
});
print_r($events);
编辑:从今天开始和持续事件被下令‘吗?他们从今天开始’,”被?他们目前正在进行中”,‘起始日期’,‘END_DATE’
$today = date('Ymd',strtotime('today'));
//function compares two elements to decide their position relative to each other
usort($events,function($a,$b) use($today){
//check if event is currently ongoing
//starts or ends today or today is between start and end
$ongoing = function($event) use($today){
//this is the simplest form I could think of for the check
return (strnatcmp($event['start_date'],$today) < 1 && strnatcmp($today,$event['end_date']) < 1);
};
//only one of them is ongoing
if($ongoing($a) && !$ongoing($b)){return -1;}
else if($ongoing($b) && !$ongoing($a)){return 1;}
//both are ongoing
else if($ongoing($a) && $ongoing($b)){
//one starts today
if($a['start_date'] === $today && $b['start_date'] != $today){return -1;}
else if($b['start_date'] === $today && $a['start_date'] != $today){return 1;}
//both start today
else if($a['start_date'] === $today && $b['start_date'] === $today){
return strnatcmp($a['end_date'],$b['end_date']);
}
//none starts today
else {
if($a['start_date'] != $b['start_date']){
//start_dates differ, order by them
return strnatcmp($a['start_date'],$b['start_date']);
} else {
//start dates are the same, order by end_date
return strnatcmp($a['end_date'],$b['end_date']);
}
}
} else {
//neither are ongoing, compare start_date
if($a['start_date'] != $b['start_date']){
//start_dates differ, order by them
return strnatcmp($a['start_date'],$b['start_date']);
} else {
//start dates are the same, order by end_date
return strnatcmp($a['end_date'],$b['end_date']);
}
}
});
print_r($events);
和最后一项可以简化为:
$today = date('Ymd',strtotime('today'));
usort($events,function($a,$b) use($today){
$ongoing = function($event) use($today){
return (strnatcmp($event['start_date'],$today) < 1 && strnatcmp($today,$event['end_date']) < 1);
};
if($ongoing($a)){
if(!$ongoing($b)){return -1;}
else {
if($a['start_date'] === $today && $b['start_date'] != $today){return -1;}
else if($b['start_date'] === $today && $a['start_date'] != $today){return 1;}
}
} else if($ongoing($b)){return 1;}
if($a['start_date'] != $b['start_date']){
return strnatcmp($a['start_date'],$b['start_date']);
} else {
return strnatcmp($a['end_date'],$b['end_date']);
}
});
print_r($events);
感谢彼得,是有道理的 - 没有想到只是彼此适用每一种。 – addedlovely
@addedlovely请记住,每次排序都会撤销之前的任何排序。您是否希望只有一个分拣机,这三个规则按降序排列? –
@addedlovely我添加了所有规则组合的解决方案,并考虑了当前正在进行的事件。 –
你想三种类型的排序或在一个所有这些规则? –