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I have the following code:
$now = date("Y-m-d H:m:s"); $date = date("Y-m-d H:m:s", strtotime('-24 hours', $now)); However, now it gives me this error:
A non well formed numeric value encountered in... why is this?
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php $starttime = strtotime('2017-05-11 21:00:00'); $starttimeformat = date('Y-m-d H:i:s', $starttime); echo "Current Time:"; echo $starttimeformat; echo '<br/>'; echo '<br/>'; $onedayadedtime_format = date('Y-m-d H:i:s', strtotime('+24 hours', $starttime)); echo "End Time after adding 24 hours:"; echo $ ...
To subtract hours from a given timestamp, we are going to use the datetime and timedelta classes of the datetime module. Step 1: If the given timestamp is in a string format, then we need to convert it to the datetime object. For that we can use the datetime. strptime() function.
The date_sub() function subtracts some days, months, years, hours, minutes, and seconds from a date.
The strtotime() function parses an English textual datetime into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 GMT). Note: If the year is specified in a two-digit format, values between 0-69 are mapped to 2000-2069 and values between 70-100 are mapped to 1970-2000.
$date = (new \DateTime())->modify('-24 hours'); or
$date = (new \DateTime())->modify('-1 day'); (The latter takes into account this comment as it is a valid point.)
Should work fine for you here. See http://PHP.net/datetime
$date will be an instance of DateTime, a real DateTime object.
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