对不起，如果问题的标题不清楚，我无法更准确地总结它。
这是问题所在：
首先，我有一个这种格式的数组：
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
)
然后一些成员未设置，所以之后我们留下如下内容：
Array (
[0] => 09:00
[1] => 10:00
[6] => 15:00
[7] => 16:00
[8] => 17:00
[9] => 18:00
)
如您所见，数组表示时隙。现在，我需要做的是消除所有短于 3 小时的时间段。所以我需要遍历数组，只要原始数组的成员少于 3 个，也就把它们拿出来。所以在上面的例子中，由于 09：00 和 10：00 后面没有 11：00，我需要把它们拿出来，留下：
Array (
[6] => 15:00
[7] => 16:00
[8] => 17:00
[9] => 18:00
)
我该如何实现此目的？从逻辑上讲，我认为检查 3 个连续索引可能最简单，而不是检查实际时间，但我愿意接受任何建议。
我已经自己解决了这个问题，我把它变成了通用的，所以它可以在任何持续时间内工作，而不仅仅是 3 小时。
$dur=3; //could be anything
foreach($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]);
}
$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);
试试这个：
<?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--;
}
}
?>
希望这有帮助
$a = Array(
0 => "09:00",
1 => "10:00",
6 => "15:00",
7 => "16:00",
8 => "17:00",
9 => "18:00",
11 => "20:00",
);
foreach ($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]);
}