Reputation: 780
I was trying to display free timeslots.
Given the data below, basically we need to find a way to display not booked timeslots during opening hours. Seems very easy to do as a human, but programming it...honestly I'm just going crazy :D
// open hours --------++++--++++----- [[08:00,12:00], [14:00,18:00]]
// booked slots ---------++----+------- [[09:00,11:00], [15:00,16:00]]
// expected --------+--+--+-++----- [[08,09], [11,12], [14,15], [16,18]]
Just for clarity I omitted minutes, which in real program will be present.
I have prepared a fiddle to start with: https://ideone.com/Z9pPi3
<?php
$opening_hours = [['08:00','12:00'], ['14:00','18:00']];
$occupied_slots = [['09:30','14:00'], ['15:10','16:35']];
$expected_result = [['08:00','09:30'], ['11:00','12:00'], ['14:00','15:10'], ['16:35','18:00']];
$valid_timeslots = [];
# - - - - - - - - helper functions
function timestring_to_time($hh_mm) {
return (int) strtotime("1970-01-01 $hh_mm");
}
function timestring_diff($hh_mm_start, $hh_mm_end) {
return abs(timestring_to_time($hh_mm_end) - timestring_to_time($hh_mm_start));
}
# find empty timeslots during opening hours given occupied slots
# H E R E G O E S T H E M A G I C
var_dump($valid_timeslots);
I tried to solve with if/else method but it's not working actually...there is a need of some kind recursion function.
Upvotes: 1
Views: 612
Reputation: 92467
Here is my solution - I assume that first hour in time-interval like ['08:00','12:00'] is always smaller than second hour. Instead of using your timestring_to_time
and timestring_diff
I write my own procedures to convert time to number - timeToNum
and numToTime
(you can easily extend them to include seconds: num=3600*hour + 60*min + sec
, sec=num%60, h=floor(num/3600), min=floor((num-h*3600)/60)
):
<?php
$opening_hours = [['08:00','12:00'], ['14:00','18:00']];
$occupied_slots = [['09:30','11:00'], ['15:10','16:35']];
$expected_result = [['08:00','09:30'], ['11:00','12:00'], ['14:00','15:10'], ['16:35','18:00']];
$valid_timeslots = [];
#find empty timeslots during opening hours given occupied slots
function timeToNum($time) {
preg_match('/(\d\d):(\d\d)/', $time, $matches);
return 60*$matches[1] + $matches[2];
}
function numToTime($num) {
$m = $num%60;
$h = intval($num/60) ;
return ($h>9? $h:"0".$h).":".($m>9? $m:"0".$m);
}
// substraction interval $b=[b0,b1] from interval $a=[a0,a1]
function sub($a,$b)
{
// case A: $b inside $a
if($a[0]<=$b[0] and $a[1]>=$b[1]) return [ [$a[0],$b[0]], [$b[1],$a[1]] ];
// case B: $b is outside $a
if($b[1]<=$a[0] or $b[0]>=$a[1]) return [ [$a[0],$a[1]] ];
// case C: $a inside $b
if($b[0]<=$a[0] and $b[1]>=$a[1]) return [[0,0]]; // "empty interval"
// case D: left end of $b is outside $a
if($b[0]<=$a[0] and $b[1]<=$a[1]) return [[$b[1],$a[1]]];
// case E: right end of $b is outside $a
if($b[1]>=$a[1] and $b[0]>=$a[0]) return [[$a[0],$b[0]]];
}
// flat array and change numbers to time and remove empty (zero length) interwals e.g. [100,100]
// [[ [167,345] ], [ [433,644], [789,900] ]] to [ ["07:00","07:30"], ["08:00","08:30"], ["09:00","09:30"] ]
// (number values are not correct in this example)
function flatAndClean($interwals) {
$result = [];
foreach($interwals as $inter) {
foreach($inter as $i) {
if($i[0]!=$i[1]) {
//$result[] = $i;
$result[] = [numToTime($i[0]), numToTime($i[1])];
}
}
}
return $result;
}
// calculate new_opening_hours = old_opening_hours - occupied_slot
function cutOpeningHours($op_h, $occ_slot) {
foreach($op_h as $oh) {
$ohn = [timeToNum($oh[0]), timeToNum($oh[1])];
$osn = [timeToNum($occ_slot[0]), timeToNum($occ_slot[1])];
$subsn[] = sub($ohn, $osn);
}
return $subsn;
}
$oph = $opening_hours;
foreach($occupied_slots as $os) {
$oph = flatAndClean(cutOpeningHours($oph, $os ));
}
$valid_timeslots = $oph;
var_dump(json_encode(["result"=>$valid_timeslots]));
Working example HERE.
Calculate $new_opening_hours
from $old_opening_hours
by substract one occupied slot from it. And repeat that operation for every slot (each time getting new oppening hours array)
To make substraction of two interwals I :
08*60+30 = 510
sub
function) - as example [500,800]-[600,700] = [ [500,600], [700,800]],You can improve a little this solution by not conver time-number on each iteration but do it for each input data at the beginning and conver number-time for each output data at the end. And probably you can reduce number of conditions in sub
function - however current version is very clear.
Upvotes: 2