我有一系列的时间字符串,比如"8:00am","8:15am","8:30am“,我也有这些时间的时间戳作为数组的键,因为时间是值,所以数组应该是这样的:
注意:在本例中,时间戳是虚拟的
array(
144441415=>"8:00am",
1444784744=>8:30am
...
.....
);
我想知道小时是否整小时,例如“8:00 or”、“9:00 or”是整小时,但“8:45 or”或“9:30 or”则不是整小时。我想从上面提到的数组中过滤整个小时。谢谢。
发布于 2015-11-10 03:53:59
谢谢你,Rudie,我能够从几秒钟中得到整个小时,谢谢你的提示,所以我做了这样的事情:
//Loop through each hour of the day by getting both key and value
foreach( $hours_of_day $timestamp=>$time_string )
{
//If timestamp mod by number of seconds in hours is zero this means whole hour
if( ($timestamp%3600) == 0 )
{
echo "Whole hour = $timestring<br />";
}
}
发布于 2015-11-10 03:47:59
你可以这样做:
$times = array(
144441415=>"8:00am",
1444784744=>"8:30am",
1444784745=>"8:45am",
1444784746=>"9:00am",
1444784747=>"10:00am",
);
foreach ($times as $time) {
if (date("H:00:00", strtotime($time)) == date("H:i:00", strtotime($time))) {
echo '<br>Whole Hour: '.$time;
}
else {
echo '<br>Not Whole Hour: '. $time;
}
}
发布于 2018-07-23 03:21:44
对于那些可能遇到这个问题的人来说,实现这一点的最佳方法是:
define("BR", "<br />");
$data = array(
144441415=>"8:00am",
1444784744=>"8:30am"
);
$tmp_dto;
$tmp_mins;
foreach($data as $hour){
$tmp_dto = DateTime::createFromFormat("H:iA", $hour);
$tmp_mins = $tmp_dto->format("i");
if($tmp_mins == "00"){
echo "ROUND".BR;
}
else{
echo "NOT ROUND".BR;
}
}
输出
ROUND
NOT ROUND
https://stackoverflow.com/questions/33616800
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