Check PHP array for consecutive indexes - php

Question

Sorry if the title of the question is unclear, I couldn't sum it up more precisely.
This is the issue:
To begin with, I have an array in this format:
Array (
[0] => 09:00
[1] => 10:00
[2] => 11:00
[3] => 12:00
[4] => 13:00
[5] => 14:00
[6] => 15:00
[7] => 16:00
[8] => 17:00
[9] => 18:00
)
Then some of the members are unset, so after that we're left with something like:
Array (
[0] => 09:00
[1] => 10:00
[6] => 15:00
[7] => 16:00
[8] => 17:00
[9] => 18:00
)
As you see, the array represents time slots. Now, what I need to do is eliminate all time slots shorter than 3 hours. So I need to go through the array and, wherever there are less than 3 members of the original array present, take them out too. So in the example above, since 09:00 and 10:00 are not followed by 11:00, I need to take them out and be left with:
Array (
[6] => 15:00
[7] => 16:00
[8] => 17:00
[9] => 18:00
)
How do I accomplish this? Logically, I think it might be easiest to check for 3 consecutive indexes, rather then checking the actual times but I'm open to any suggestions.

Answer 1

I've solved the problem on my own, and I made it generic so it would work for any duration, not just 3 hours.
$dur=3; //could be anything
foreach($work_times as $member){
$key=array_search($member,$work_times);
$a_ok=0;
for($options=0;$options<$dur;$options++){
$thisone=1;
for($try=$key-$options;$try<$key-$options+$dur;$try++){
if(!array_key_exists($try,$work_times))
$thisone=0;
}
if($thisone==1)
$a_ok=1;
}
if($a_ok==0)
unset($work_times[$key]);
}

Answer 2

$arr = array(
0 => '09:00',
1 => '10:00',
6 => '15:00',
7 => '16:00',
8 => '17:00',
9 => '18:00'
);
// for some testing
$arr += array(12 => '19:00', 13 => '20:00', 14 => '21:00');
$arr += array(16 => '22:00', 17 => '23:00');
$dontRemove = array();
foreach($arr as $key => $val) {
// if the next 2 keys are set
if (isset($arr[$key+1]) && isset($arr[$key+2])) {
$dontRemove[] = $key;
$dontRemove[] = $key+1;
$dontRemove[] = $key+2;
}
}
// combine and diff the keys to get the keys which should be actually removed
$remove = array_diff(array_keys($arr), array_unique($dontRemove));
foreach($remove as $key) {
unset($arr[$key]);
}
print_r($arr);

Answer 3

Try this:
<?php
function check() {
global $array;
$tmpArr = array_keys( $array );
$val1 = $tmpArr[0];
$val2 = $tmpArr[1];
$val3 = $tmpArr[2];
if( ( ++$val1 == $val2 ) && ( ++$val2 == $val3 ) ) {
// continuous
} else {
// not continuous, remove it
unset( $array[$tmpArr[0]] );
}
}
$array = array(
'0' => '09:00',
'1'=> '10:00',
'6' => '15:00',
'7'=> '16:00',
'8' => '17:00',
'9' => '18:00'
);
$total = count( $array );
$ctotal = 0;
while( $ctotal < $total ) {
if( count( $array ) <= 2 ) {
// this array has 2 elements left, which obviously
// nullifies the 3 continuous element check
$array = array();
break;
} else {
//check the array backwards
check();
$total--;
}
}
?>
Hope this helps

Answer 4

$a = Array(
0 => "09:00",
1 => "10:00",
6 => "15:00",
7 => "16:00",
8 => "17:00",
9 => "18:00",
11 => "20:00",
);
foreach ($a as $k => $v) {
// previous or next two time slots exist
$consecutive = (isset($a[$k-1]) or
(isset($a[$k+1]) and isset($a[$k+2])));
if (!$consecutive)
unset($a[$k]);
}