我需要检查当前时间是否在范围
之间 if((strtotime($s)<=strtotime($myTime)) && (strtotime($e)>=strtotime($myTime)))
{$open =1;}
if($e=="00:00")
{ if((strtotime($s)<=strtotime($myTime))&& (strtotime($myTime)>=strtotime("00:00")))
{$open =1;}}
必须检查所有案例
var app = angular.module("myApp", []);
app.controller("myCtrl", function($scope) {
});这不适用于我出错的地方
答案 0 :(得分：1)
这样的事可能有用。我还没有真正测试过它。
$currentTime = time();
function in_range($what, $min, $max) {
return $what >= $min && $what <= $max;
}
var_dump(in_range($currentTime, strtotime('08:35'), strtotime('15:30'));
// etc.
答案 1 :(得分：1)
使用DateTime对象进行比较。见http://php.net/manual/en/class.datetime.php
$cases = array(
array(
'start' => new DateTime("10:00:00"),
'end' => new DateTime("12:00:00"),
),
array(
'start' => new DateTime("10:00:00"),
'end' => new DateTime("23:59:00"),
),
array(
'start' => new DateTime("10:00:00"),
'end' => new DateTime("00:00:00"),
),
array(
'start' => new DateTime("23:59:00"),
'end' => new DateTime("05:00:00"),
),
);
//adjust dates
$cases[2]['end']->modify('+1 day');
$cases[3]['start']->modify('-1 day');
$now = new DateTime();
foreach ($cases as $key => $value) {
if ($now > $value['start'] && $now < $value['end']) {
echo 'case ' . $key . ' ok' . PHP_EOL;
}
else {
echo 'case ' . $key . ' out of range' . PHP_EOL;
}
}
答案 2 :(得分：0)
对吧，你身边一团糟。你想看看这是否达到你的预期产量？似乎正在进行我的本地测试。公平的代码思想，原来是一些需要考虑的边缘情况。
它的潜在部分可以重构和改进，但如果它按预期工作，我会把它留给你。
<?php
$cases = [
[['start' => '09:00', 'end' => '01:00']],
[['start' => '10:00', 'end' => '23:59']],
[['start' => '10:00', 'end' => '00:00']],
[['start' => '23:59', 'end' => '05:00']],
[['start' => '10:00', 'end' => '15:00'], ['start' => '17:30', 'end' => '02:00']],
[['start' => '10:00', 'end' => '15:00'], ['start' => '17:30', 'end' => '00:00']],
];
$cases = setUpCases($cases);
foreach ($cases as $case) {
$withinTimeRange = isTimeWithinCase((new DateTime()), $case);
if ($withinTimeRange) {
break;
}
}
// Boolean True/False whether it is within any of the ranges or not.
// i.e. True = Shop Open, False = Shop closed.
var_dump($withinTimeRange);
function setUpCases($cases) {
return array_map(function($case) {
return convertToDateTimeObjects($case);
}, $cases);
}
function convertToDateTimeObjects($case) {
$formatted = [];
foreach ($case as $index => $times) {
$s = new DateTime();
list($hour, $minute) = explode(':', $times['start']);
$s->setTime($hour, $minute, 00);
$e = new DateTime();
list($hour, $minute) = explode(':', $times['end']);
$e->setTime($hour, $minute, 00);
if ($e < $s) $e->modify('+1 day');
$formatted[] = ['start' => $s, 'end' => $e];
}
return $formatted;
}
function isTimeWithinCase($time, $case) {
foreach ($case as $timerange) {
return ($time > $timerange['start'] && $time < $timerange['end']);
}
}
答案 3 :(得分：0)
我一直在仔细考虑这项任务。我想出了计算当前时间是否在2倍之间所需的逻辑。有两种情况可以计算：
这意味着结束时间必须是第二天。这意味着如果您当前时间小于结束时间OR大于开始时间，则您的时间介于两次之间。
在这种情况下，开始时间和结束时间是同一天。为此，需要的逻辑是检查您的时间是否大于开始时间AND小于结束时间。如果该陈述为真，则您的时间介于两次之间。
我已经计算了以下函数：
function checktime($arr,$time) {
list($h,$m)=explode(":",$time);
if(!is_array($arr[0])) {
$r1=explode(":",$arr[0]);
$r2=explode(":",$arr[1]);
if($r1[0]>$r2[0]) {
if(($h>$r1[0] || ($h==$r1[0] && $m>=$r1[1])) || ($h<$r2[0] || ($h==$r2[0] && $m<=$r2[1]))) return true;
}
else {
if(($h>$r1[0] || ($h==$r1[0] && $m>=$r1[1])) && ($h<$r2[0] || ($h==$r2[0] && $m<=$r2[1]))) return true;
}
}
}
以下是一些测试，用于测试你给我的案例的限制：
$arr=array("10:00","01:00");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//True
echo (checktime($arr,"00:00")?"True":"False");//True
$arr=array("10:00","23:59");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//True
echo (checktime($arr,"00:00")?"True":"False");//False
$arr=array("10:00","00:00");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//True
echo (checktime($arr,"00:00")?"True":"False");//True
$arr=array("23:59","05:00");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//False
echo (checktime($arr,"00:00")?"True":"False");//True
$arr=array("08:35","15:30");
echo (checktime($arr,"23:59")?"True":"False");//False
echo (checktime($arr,"12:00")?"True":"False");//True
echo (checktime($arr,"00:00")?"True":"False");//False
$arr=array("17:30","02:00");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//False
echo (checktime($arr,"00:00")?"True":"False");//True
$arr=array("10:00","15:00");
echo (checktime($arr,"23:59")?"True":"False");//False
echo (checktime($arr,"12:00")?"True":"False");//True
echo (checktime($arr,"00:00")?"True":"False");//False
$arr=array("17:00","00:00");
echo (checktime($arr,"23:59")?"True":"False");//True
echo (checktime($arr,"12:00")?"True":"False");//False
echo (checktime($arr,"00:00")?"True":"False");//True
答案 4 :(得分：0)
这适用于所有案例
$let=array("01:00","02:00","03:00","05:00","04:00");
if((strtotime($s)<=strtotime($myTime)) && (!strcmp($e,"00:00")))
{ $open=1;$flag=1;}
if((strtotime($s)<=strtotime($myTime)) && (!strcmp($e,"23:59")))
{ $open=1;$flag=1;}
if(in_array($e,$let))
{
if(strtotime($s)<=strtotime($myTime))
{ $open=1;$flag=1;}
else
if(strtotime($e)>=strtotime($myTime))
{ $open=1;$flag=1;}
}
if(!$flag)
{if((strtotime($s)<=strtotime($myTime)) && (strtotime($e)>=strtotime($myTime)))
{ $open=1;} }