I am trying to check if the user has been on the site longer then 6 months, if they have to prompt a message.
I cannot get this to work, it keeps thinking 6 weeks has passed even when the date is moved to today.
Currently my code.
$created_at = "2014-12-01 16:58:23";
$sixweek = 604800 * 6;
if(strtotime($created_at) < time() + ($sixweek))
{
$data['needsnewimage'] = 1;
die('6 Weeks has passed ');
}else{
$data['needsnewimage'] = 0;
die('6 Weeks has not passed');
}
Any ideas?
That's not going to work. Consider this: someone signs up for your site TODAY:
$created_at = '2014-12-01 08:00:00'; // "today"
if ($created_at < $today + $sixweek) ...
becomes
if ($today < $today + $sixweek)
if (0 < $sixweek)
and will always be true.
You want
if ($created_at > (time() - $sixweek))
^---------^
Note the changed math
DateTime will make this simple, something like:-
$created_at = new \DateTime("2014-12-01 16:58:23");
$sixWeeks = new \DateInterval('P6W');
if($created_at->add($sixWeeks) < new \DateTime()){
echo "Six weeks has passed!\n";
} else echo "Not yet!\n";
See it working
Reference
Related
I have following code that add 2 days to a given date.
$myDate = 2018-07-28 11:00:00; // the date is picked from db
$penaltyDays = 2;
$date1 = new DateTime($myDate);
$date1->add(new DateInterval("P{$penaltyDays}D")); // add N days
$now = new DateTime();
$interval = $date1->diff($now); // get difference
$days = $interval->d; // difference in days
I want value of $days must be 0 after passing exactly 48 hours. If 3 days are passed the value of $days should be -1.
I will also appreciate if someone tell me efficient/proper way to get the result.
To make an efficient code according to your specification then I would rather use strtotime than DateTime.
This code checks if the current time is larger than the database time + two (or three) days in seconds.
$myDate = "2018-07-28 11:00:00";
$unix = strtotime($myDate);
if(time() > ($unix + 86400*3)){
$days = -1;
}else if(time() > ($unix + 86400*2)){
$days = 0;
}else{
$days = "something else";
}
Echo $days;
https://3v4l.org/d6Q15
In my code I pretty much send a token to the database with the
date('Y-m-d H:i:s');
function. But I am checking to see if the timestamp I sent if greater than 60 seconds if so i echo yes and else no. I keep getting no and I know its more than a minute because i time it. I've seen post about this but they're using specific dates and I am just going on when a user submits a form and checking if the token is 60 seconds old. Here is my code
php
<?php
require_once('db_login.php');
$stmtToken = $handler->prepare("SELECT * FROM email_token");
$stmtToken->execute();
$rowToken = $stmtToken->fetch();
$date = $rowToken['time_stamp'];
if($date > time() + 60) {
echo 'yes';
} else {
echo 'no';
}
?>
You can also play with dates in different manners. All four lines here are equivalent:
$now = (new \DateTime(date('Y-m-d H:i:s')))->getTimestamp();
$now = (new \DateTime('now'))->getTimestamp();
$now = (new \DateTime())->getTimestamp();
$now = time();
And then you can compare in this manner:
$tokenExpirationTimestamp = (new \DateTime($date))
->modify('+60 seconds')
->getTimestamp();
$isTokenExpired = $tokenExpirationTimestamp < time();
if ($isTokenExpired) {
// ...
}
When you compare times and dates you can either use datetime or strtotime.
Using strings will not work as expected in all cases.
In comments you mentioned how you want to compare, and you need to add the 60 seconds to the "date", not the time().
if(strtotime($date) + 60 < time()) {
I am practicing with dates in php. I a bit of a newbie so bear my ignorance
I am trying to see when a time is before noon.
So I have a variable coming in with this format 2014-03-07 13:28:00.000
I get the time like this
$submissonTime = date('H:i:s', strtotime($value['job_submission_date']));
then I want to set another variable as $noon and i am doing this:
$noon = date('H:i:s', '12:00:00.000');
However the value of noon is 12:00:12
what i want to do is basically:
if($submissionTime <= $noon){
//do my stuff
}
NB I want to enter the if statement when even when it is 12:00:00 and stop entering when it is 12:00:01
Any help please?
Try
$noon = date('Y-m-d 12:00:00'); // today noon with date
$submissonTime = date('Y-m-d H:i:s', strtotime($value['job_submission_date']));
if(strtotime($submissonTime) <= strtotime($noon)){
//do my stuff
}
if you want to compare only time use both format
$noon = date('12:00:00');
$submissonTime = date('H:i:s', strtotime($value['job_submission_date']));
if (date("A") == "AM")
{
// AM-Code
} else {
// PM-Code
}
Why don't you go with only one string of code getting the hour?
$Hour = date("G"); //24-hour format of an hour without leading zeros
if($Hour < 12) {
// do the code
}
Or in your case
$Hour = date("G", strtotime($value['job_submission_date']));
update
If you need 12:00:00 and not 12:00:01 and later on, you will need to define minutes and seconds:
$Hour = date("G"); //24-hour format of an hour without leading zeros
$Minute = intval(date("i")); // will give minutes without leading zeroes
$Second = intval(date("s"));
if(($Hour < 12) || ($Hour == 12 && $Minute == 0 && Second == 0)) {
// do the code
}
I need to find date x such that it is n working days prior to date y.
I could use something like date("Y-m-d",$def_date." -5 days");, but in that case it wont take into consideration the weekend or off-date. Let's assume my working days would be Monday to Saturday, any idea how I can accomplish this?
Try this
<?php
function businessdays($begin, $end) {
$rbegin = is_string($begin) ? strtotime(strval($begin)) : $begin;
$rend = is_string($end) ? strtotime(strval($end)) : $end;
if ($rbegin < 0 || $rend < 0)
return 0;
$begin = workday($rbegin, TRUE);
$end = workday($rend, FALSE);
if ($end < $begin) {
$end = $begin;
$begin = $end;
}
$difftime = $end - $begin;
$diffdays = floor($difftime / (24 * 60 * 60)) + 1;
if ($diffdays < 7) {
$abegin = getdate($rbegin);
$aend = getdate($rend);
if ($diffdays == 1 && ($astart['wday'] == 0 || $astart['wday'] == 6) && ($aend['wday'] == 0 || $aend['wday'] == 6))
return 0;
$abegin = getdate($begin);
$aend = getdate($end);
$weekends = ($aend['wday'] < $abegin['wday']) ? 1 : 0;
} else
$weekends = floor($diffdays / 7);
return $diffdays - ($weekends * 2);
}
function workday($date, $begindate = TRUE) {
$adate = getdate($date);
$day = 24 * 60 * 60;
if ($adate['wday'] == 0) // Sunday
$date += $begindate ? $day : -($day * 2);
return $date;
}
$def_date="";//define your date here
$preDay='5 days';//no of previous days
date_sub($date, date_interval_create_from_date_string($preDay));
echo businessdays($date, $def_date); //date prior to another date
?>
Modified from PHP.net
Thanks for the help guys, but to solve this particular problem I wrote a simple code:
$sh_padding = 5; //No of working days to count backwards
$temp_sh_padding = 1; //A temporary holder
$end_stamp = strtotime(date("Y-m-d", strtotime($date_format)) . " -1 day"); //The date(timestamp) from which to count backwards
$start_stamp = $end_stamp; //start from same as end day
while($temp_sh_padding<$sh_padding)
{
$sh_day = date('w',$start_stamp);
if($sh_day==0){ //Skip if sunday
}
else
{
$temp_sh_padding++;
}
$start_stamp = strtotime(date("Y-m-d",$start_stamp)." -1 day");
}
$sh_st_dte = date("Y-m-d",$start_stamp); //The required start day
A quick bit of googling got me to this page, which includes a function for calculating the number of working days between two dates.
It should be fairly trivial to adjust that concept to suit your needs.
Your problem, however, is that the concept of "working days" being monday to friday is not universal. If your software is only ever being used in-house, then it's okay to make some assumptions, but if it's intended for use by third parties, then you can't assume that they'll have the same working week as you.
In addition, public holidays will throw a big spanner in the works, by removing arbitrary dates from various working weeks throughout the year.
If you want to cater for these, then the only sensible way of doing it is to store the dates of the year in a calendar (ie a big array), and mark them individually as working or non-working days. And if you're going to do that, then you may as well use the same mechanism for weekends too.
The down-side, of course, is that this would need to be kept up-to-date. But for weekends, at least, that would be trivial (loop through the calendar in advance and mark weekend days where date('w')==0 or date('w')==6).
Any idea in PHP how to get a unix timestamp of the next occurring specified hour and minute?
Thank you
Calling strtotime("4pm") will give you the time for today at 4pm. If it's already past 4pm, you can just add 60*60*24 to the given timestamp
// Example blatantly copied from AndrewR, but it uses strtotime
$nextfourpm = strtotime("4pm");
if ($nextfourpm < time()) {
$nextfourpm += 60*60*24;
}
You could do something like this.
$nextTime = mktime(16, 0, 0);
if($nextTime < time()){
$nextTime += 86400;
}
Now is 15:20
You want tomorrow at 16:22
BEST way for you is to create function with posting hour+ and minutes+ you want to go beyond - pure difference in time). By doing this way PHP cannot make any mistakes. And PHP can make mistakes while working with dates and stuff.
//stuff to post..
$hour = 25;
$minute = 2;
function makemydate($hour,$minute)
{
if($hour > 0)
{
$hour = ($hour*3600);
}
if($minute > 0)
{
$minute = ($minute*60);
}
$add = ($hour+$minute);
$now = time();
$unixtimestamp = ($now+$add);
return $unixtimestamp;
}
echo makemydate;
There you go its a unix timestamp.. You need to specify time whatsoever :P