以下功能输出小时数:0是时间是<1小时还是分钟:0分钟时<1。
如何只显示不为零的变量?
谢谢。
function time_difference($endtime){
$hours =date("G",$endtime);
$mins =date("i",$endtime);
$secs =date("s",$endtime);
$diff="'hours': ".$hours.",'mins': ".$mins.",'sec': ".$secs;
return $diff;
}
$end_time =strtotime("+7 hours") - strtotime($entry->pubDate);
$difference = time_difference($end_time);
echo $difference;
答案 0 :(得分:3)
另一种可能的方法:
function time_difference($endtime){
$times=array(
'hours' => date("G",$endtime),
'mins' => date("i",$endtime),
'secs' => date("s",$endtime),
);
//added a "just a moment ago" feature for you
if (intval($times['hours'], 10) == 0
&& intval($times['mins'], 10) == 0) {
return "just a moment ago";
}
$diff='';
foreach ($times as $k=>$v) {
$diff.=empty($diff) ? '' : ',';
$diff.=intval($v, 10) == 0 ? '' : "'$k':$v";
}
return $diff;
}
答案 1 :(得分:2)
使用?操作
$diff=($hours > 0) ? "'hours': ".$hours : "";
$diff=$diff.($minutes > 0) ? etc...
答案 2 :(得分:1)
对于较大的时间范围,您最好使用数学而不是date()
:
function time_difference($endtime){
// hours can get over 23 now, $endtime is in seconds
$hours = floor($endtime / 3600);
// modulo (%) already rounds down, not need to use floor()
$mins = $endtime / 60 % 60;
// the remainder of $endtime / 60 are seconds in a minute
$secs = $endtime % 60;
// this array holds the hour, minute and seconds if greater than 0
$diff = array();
if ($hours) $diff[] = "'hours': $hours";
if ($mins) $diff[] = "'mins': $mins";
if ($secs) $diff[] = "'sec': $secs";
// join the values with a comma
$diff = implode(',', $diff);
if (!$diff) { // hours, mins and secs are zero
$diff = "just a moment ago";
}
return $diff;
}
以下功能只返回0-23范围内的小时数。如果时间超过一天,则小时变为零:
function time_difference($endtime){
$hours = (int)date("G",$endtime);
$mins = (int)date("i",$endtime);
$secs = (int)date("s",$endtime);
// this array holds the hour, minute and seconds if greater than 0
$diff = array();
if ($hours) $diff[] = "'hours': $hours";
if ($mins) $diff[] = "'mins': $mins";
if ($secs) $diff[] = "'sec': $secs";
// join the values with a comma
$diff = implode(',', $diff);
if (!$diff) { // hours, mins and secs are zero
$diff = "just a moment ago";
}
return $diff;
}
将(int)
返回的字符串转换为字符串需要 date()
。使用此"01"
变为1,"00"
变为“0”。