The remaining time refers to how many minutes or days it has been since an article was published. In many blog forums, you can see that the article was published 1 day ago. Below I will introduce two examples
Combine the two The string in date format is converted into a unix timestamp, and then subtracted to obtain the timestamp difference. Finally, judge the remaining time and generate a time format similar to (published 2 hours, 30 minutes and 20 seconds ago)
The code is as follows
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public function gettime($time_s,$time_n){
$time_s = strtotime($time_s);
$time_n = strtotime($time_n);
$strtime = '';
$time = $time_n-$time_s;
if($time >= 86400){
return $strtime = date('Y-m-d H:i:s',$time_s);
}
if($time >= 3600){
$strtime .= intval($time/3600).'小时';
$time = $time % 3600;
}else{
$strtime .= '';
}
if($time >= 60){
$strtime .= intval($time/60).'分钟';
$time = $time % 60;
}else{
$strtime .= '';
}
if($time > 0){
$strtime .= intval($time).'秒前';
}else{
$strtime = "时间错误";
}
return $strtime;
}
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Copy code
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public function gettime($time_s,$time_n){
$time_s = strtotime($time_s);
$time_n = strtotime($time_n);
$strtime = '';
$time = $time_n-$time_s;
if($time >= 86400){
return $strtime = date('Y-m-d H:i:s',$time_s);
}
if($time >= 3600){
$strtime .= intval($time/3600).'hour';
$time = $time % 3600;
}else{
$strtime .= '';
}
if($time >= 60){
$strtime .= intval($time/60).'minutes';
$time = $time % 60;
}else{
$strtime .= '';
}
if($time > 0){
$strtime .= intval($time).'seconds ago';
}else{
$strtime = "Wrong time";
}
return $strtime;
}
First determine whether the value you want to subtract is greater than the number of seconds in a day, 86400 seconds. If it is greater, return the time queried from the original database
Then determine whether it is within 1 hour to one day, that is, 3600 seconds-86400 seconds. If it is within, return X hours. After getting the result, you need to use the remainder method to remove the hour part of the time, use % to take the remainder |
Then determine whether it is within 1 minute to one hour, that is, 60 seconds - 3600 seconds. If it is within, return X minutes. After getting the result, you need to use the remainder method to remove the minute part of the time, use % to take the remainder
Finally determine whether it is within 1 minute, that is, 0 seconds - 60 seconds. If it is within 1 minute, return X minutes and seconds
Note: The results obtained above are all connected using .=. In this way, we finally get an overall time.
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http: //www.bkjia.com/PHPjc/632724.htmlThe remaining time refers to the minutes or days since an article was published. This is used in many blogs. Everyone in the forum has seen the article posted 1 day ago. Below I will introduce two examples...
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