自开始和结束时间在php中的20分钟时间槽

Question

Ex ->

 Start time = 2015-10-21 09:00:00
 end time  = 2015-10-21 19:45:00

因此，根据我的示例-开始时间是9 am to 7-45pm

所以我想要20分钟的时间。在php类

9:00 - 9:20 , 9-20 to 9-40 etc。但对于ex 12:00-12:20 , 2:40 to 2:60预订，那么它将不包括在这个列表中。

我想安排20分钟的时间。是的，但是它也会过滤alr。

使用php中的时间差，我有开始时间和结束时间。

然后我要检查从开始到结束的时间可能有多少20分钟。

=>它也会检查某个时段是否已经预订，然后不考虑它。

IN SHOT 1 to 2 or 3-5 is break time then it will not consider. 

Answer 1

这能满足你的需要。但是，为了便于计算，应该为数组元素保留特定的格式。

<?php
$start_time = '2015-10-21 09:00:00';  //start time as string
$end_time = '2015-10-21 19:45:00';  //end time as string
$booked = array('12:20-12:40','13:00-13:20');    //booked slots as arrays
$start = DateTime::createFromFormat('Y-m-d H:i:s',$start_time);  //create date time objects
$end = DateTime::createFromFormat('Y-m-d H:i:s',$end_time);  //create date time objects
$count = 0;  //number of slots
$out = array();   //array of slots 
for($i = $start; $i<$end;)  //for loop 
{
$avoid = false;   //booked slot?
$time1 = $i->format('H:i');   //take hour and minute
$i->modify("+20 minutes");      //add 20 minutes
$time2 = $i->format('H:i');     //take hour and minute
$slot = $time1."-".$time2;      //create a format 12:40-13:00 etc
    for($k=0;$k<sizeof($booked);$k++)  //if booked hour
    {
    if($booked[$k] == $slot)  //check
    $avoid = true;   //yes. booked
    }
if(!$avoid && $i<$end)  //if not booked and less than end time
{
$count++;           //add count
$slots = ['start'=>$time1, 'stop'=>$time2];         //add count
array_push($out,$slots); //add slot to array
}
}
var_dump($out);   //array out
echo $count ." of slots available";

如果使用已预订的日期时间作为数组，请使用下面的代码。

<?php
$start_time = '2015-10-21 09:00:00';  //start time as string
$end_time = '2015-10-21 19:45:00';  //end time as string
$booked = ['2015-10-21 12:20:00','2015-10-21 12:40:00', '2015-10-21 13:00:00','2015-10-21 13:20:00'];
     //booked slots as arrays
$start = DateTime::createFromFormat('Y-m-d H:i:s',$start_time);  //create date time objects
$end = DateTime::createFromFormat('Y-m-d H:i:s',$end_time);  //create date time objects
$time1 = $start;
$count = 0;  //number of slots
$out = array();   //array of slots 
for($i = $start; $i<$end;)  //for loop 
{
$avoid = false; 
$t1 = date_timestamp_get($i);
$t2 = $t1+(20*60);

    for($k=0;$k<sizeof($booked);$k+=2)  //if booked hour
    {
    $st = DateTime::createFromFormat('Y-m-d H:i:s',$booked[$k]);
    $en = DateTime::createFromFormat('Y-m-d H:i:s',$booked[$k+1]);

    if( $t1 >= date_timestamp_get($st) && $t2 <= date_timestamp_get($en)  )
    $avoid = true;   //yes. booked
    }
$slots =[ $i->format('H:i'),$i->modify("+20 minutes")->format('H:i')];
if(!$avoid && $i<$end)  //if not booked and less than end time
{
$count++;  
array_push($out,$slots);  //add slot to array
}
}
var_dump($out);   //array out
echo $count ." of slots available";

Answer 2

您必须循环遍历直到最后一个单循环(小于或等于end time)，同时我们每次增加start time 20分钟，并将其添加到数据数组中。

您可以在PHP文档页面上看到有关我使用的函数的更多信息：strtotime，日期

<?php

    $start_time = strtotime('2015-10-21 09:00:00');
    $end_time = strtotime('2015-10-21 19:45:00');
    $slot = strtotime(date('Y-m-d H:i:s',$start_time) . ' +20 minutes');

    $data = [];

    for ($i=0; $slot <= $end_time; $i++) { 

        $data[$i] = [ 
            'start' => date('Y-m-d H:i:s', $start_time),
            'end' => date('Y-m-d H:i:s', $slot),
        ];

        $start_time = $slot;
        $slot = strtotime(date('Y-m-d H:i:s',$start_time) . ' +20 minutes');
    }

    print_r($data);

?>