i have to sent an email when a user register email contain a link that is become invalid after six hours
what i m doing when email is sent i update the db with field emailSentDate of type "datetime"
now i got the curent date and time and has made to the same formate as it is in db now i want to find that both these dates and time have differenc of 6 hours or not so that i can make link invalid but i donot know how to do this
my code is look like this i m using hardcoded value for db just for example
$current_date_time=date("Y-m-d h:i:s");
$current=explode(" ",$current_date_time);
$current_date=$current[0];
$current_time=$current[1];
$db_date_time="2010-07-30 13:11:50";
$db=explode(" ",$db_date_time);
$db_date=$db[0];
$db_time=$db[1];
i do not know how to proceed plz help
<?php
//$now = new DateTime(); // current date/time
$now = new DateTime("2010-07-28 01:11:50");
$ref = new DateTime("2010-07-30 05:56:40");
$diff = $now->diff($ref);
printf('%d days, %d hours, %d minutes', $diff->d, $diff->h, $diff->i);
prints 2 days, 4 hours, 44 minutes
see http://docs.php.net/datetime.diff
edit: But you could also shift the problem more to the database side, e.g. by storing the expiration date/time in the table and then do a query like
... WHERE key='7gedufgweufg' AND expires<Now()
Many rdbms have reasonable/good support for date/time arithmetic.
What you can do is convert both of your dates to Unix epoch times, that is, the equivalent number of seconds since midnight on the 31st of December 1969. From that you can easily deduce the amount of time elapsed between the two dates. To do this you can either use mktime() or strtotime()
All the best.
$hoursDiff = ( time() - strtotime("2010-07-30 13:11:50") )/(60 * 60);
I'd rather work with a timestamp: Save the value which is returned by "time()" as "savedTime" to your database (that's a timestamp in seconds). Subtract that number from "time()" when you check for your six hours.
if ((time() - savedTime) > 6 * 3600)
// more than 6h ago
or
"SELECT FROM table WHERE savedTime < " . (time() - 6 * 3600)
This might be the solution to your problem -> How to calculate the difference between two dates using PHP?
Related
I want to display content from the database with dates up to 2hours ahead of time.
Example:
2018-11-09 20:00:00.000000
2018-11-08 19:00:00.000000
2018-11-06 19:00:00.000000
2018-11-06 18:00:00.000000
Lets say the time and date is
Nov 6th at 6pm. I want the bottom two entries to be displayed and the two future dates to not show until the current time is within 2hours of that time.
My code is as follows:
$cT = strtotime($row3['MissionTime']) - strtotime("now");
if($cT <= strtotime('-2 hours')) {
echo $row3['MissionTime']."<br>";
}
I've tried several different ways but I can't seem to get this to work right. Help and tips?
The reason your code doesn't work is that strtotime returns a number of seconds since the unix epoch. When you subtract two results of strtotime you will get a number of seconds difference which is as you expect. However you cannot compare that value to strtotime('-2 hours') as the output of that will be the timestamp for 2 hours before now (which right now is 1541539906), so the test will always pass. You should just compare it to 7200 instead (I'm pretty sure based on your question description that +7200 is more appropriate than -7200). so change
if($cT <= strtotime('-2 hours')) {
to
if($cT <= 7200) {
Note that it is almost certainly better to do this in your query. Try adding a condition on your time column as something like
WHERE MissionTime <= NOW() + INTERVAL 2 HOUR
And then you won't need to check in the PHP at all.
strtotime() returns a timestamp in seconds. Subtracting two timestamps gives you a difference between those two timestamps, in seconds.
So if strtotime($row3['MissionTime']) is a timestamp that's 1.5 hours in the future, and you subtract strtotime("now") from it, you end up with a difference of 5400 seconds (60 seconds * 60 minutes * 1.5 hours).
strtotime('-2 hours') gives you the timestamp for 2 hours ago, which is currently somewhere around 1.5 billion. This is not very useful for your situation.
Here are two ways to modify your code:
$cT = strtotime($row3['MissionTime']) - strtotime("now");
if($cT <= 7200) {
echo $row3['MissionTime']."<br>";
}
If the difference between $row['MissionTime'] and now is less than 7200 seconds (60 seconds * 60 minutes * two hours), $row3['MissionTime'] is either in the past or it's within the next two hours.
Alternatively:
if(strtotime($row3['MissionTime']) <= strtotime('+2 hours')) {
echo $row3['MissionTime']."<br>";
}
Basically the same, but perhaps more readable if you're not planning to use $cT for anything else. This simply checks if $row3['MissionTime'] is earlier than whatever time it will be in +2 hours.
I only have basic PHP knowledge and I'm using PHP+Mysql and trying to check the difference in days between 2 dates; the 1st date is formatted by myself in the script as a string:
$f_entrega=$_POST['year1']."-".$_POST['month1']."-".$_POST['day1'];
The second date ($f_dock) which is the one causing the issue is taken from the mysql database which column is in DATE format. To compare the dates I do the following:
if(!empty($_POST["id"])){
$f_entrega=$_POST['f_entrega_a']."-".$_POST['f_entrega_m']."-".$_POST['f_entrega_d'];
$f_entr=$f_entrega;
$mysqli=conectar();
$resultado = $mysqli->query("SELECT id,f_dock FROM pt");
$row = $resultado->fetch_assoc();
for ($i=0;$i<count($ids);$i++){
do{
if ($ids[$i]==$row["id"]){
$f_dock=$row["f_dock"];
break;
}
} while ($row = $resultado->fetch_assoc());
$error=0;
var_dump($f_dock);
$f_dock=strtotime($f_dock);
$f_dock=date('Ymd',$f_dock);
$f_entrega=$f_entr;
$f_entrega=strtotime($f_entrega);
$f_entrega=date('Ymd',$f_entrega);
$f_dock=DateTime::createFromFormat('Ymd',$f_dock);
$f_entrega=DateTime::createFromFormat('Ymd',$f_entrega);
$dias_vendor=date_diff($f_dock,$f_entrega);
$tat=$dias_vendor->format('%R%a');
Sometimes it works correctly, but other times I get Warning: strtotime() expects parameter 1 to be string, object given in [first line] and $tat is not correctly calculated and has strange values.
I've tried different solutions like $f_dock=(string)$f_dock before but finally the convertion always fails in some cases. Thanks in advance for any tip.
The error that you are getting is because the string you are entering is not a valid string for the strtotime() function to convert.
For instance 2015-08-31 will convert just fine, as will today, tomorrow or +7 days.
For more specific help you will need to tell us what the value of $f_dock is (as Marcos says in his comment, var_dump($f_dock) will get you this).
However, on to the solution:
$date1 = strtotime($f_dock); //timestamp in seconds
$date2 = strtotime($f_entrega); //same for the second date
$difference = $date1 - $date2; //difference in seconds between the dates
$days = floor($difference/86400);
86400 is the number of seconds in a day, so find out how many seconds difference there is, then see how many days worth of seconds are in there and use floor() to round the number down. Job done.
This may be a complete noob question but here goes:
I have the following code that compares two dates for absence management. Where I expect the answer to return as 2 (the difference between start and end date) I get 1.
$start_time = new DateTime("2015-01-01 00:00:00");
$end_time = new DateTime("2015-01-02 00:00:00");
$diff = $end_time->diff($start_time);
$d = $diff->days; // 1
I have also tried using just the dates (but I need the times as some absence type are done by hours not days)
Difference is 1 because there is only one day difference between both days.
To convert the datetime into hours or minutes you should look to these links:
Convert datetime into year, month, days, hours, minutes, seconds
Difference between 2 time() values
I'm looking for an easy way to see if more than two hours has passed between two dates. I can either do this with a MySQL DATETIME value, or if needed, I can convert that to a UNIX timestamp. I just need an easy way to to compare those two dates and see if more than 2 hours has passed.
try to look into DATEDIFF function in MySQL.
A UNIX timestamp is just the number of seconds that have elapsed since 12:00AM UTC, January 1, 1970.
Two hours in seconds is 60 * 60 * 2 = 7200. So,
if($secondTimestamp - $firstTimestamp >= 7200)
{
echo '2 hours have elapsed.';
}
Since you tagged with php, you could use PHP's DateTime::diff (DateTime::diff) to get a diff between two datetime objects. I guess it depends on where in your application you are doing the comparison.
In PHP
$time = strtotime($date2) - strtotime($date1); //this will give difference in seconds between two dates
if(($time/3600) >= 2) { // 2 hours has left }
Hi I get an error when trying to get date interval using php strtotime function
the code is:
<?php
$interval = time() - strtotime('1992/08/13');
//expect to be 18
// but the output is 1988
print date('Y', $interval);
?>
any advice?
thanks
If you want to deal with date intervals in PHP I can't recommend the DateInterval class enough. I wrote a blog post on this earlier this week: Working with Date and Time in PHP
There's an example of using it there that should allow you to do what you want to do.
That is because all time() functions are seconds since epoch which is in 1970, so your out is actually 18 years since epoch. If you want it to get the difference in years you will probably have to calculate the difference yourself.
print $interval / (60*60*24*365.242199);
Are you tring to get the years elapsed rather than the actual year?
If so:
$year = 31556926;
$interval = time() - strtotime('1992/08/13');
print round($interval / $year);
$interval = time() - strtotime('1992/08/13');
These PHP functions deal with UNIX timestamps. That means the number of seconds from 1970. 01. 01. So 1992/08/13 is transformed into a timestamp (seconds). time() gives the current timestamp (seconds). You subtract the former from the latter, and you get the amount of seconds between those two dates. This is not a date itself, just an interval.
If you want to get the year, do something like echo $interval/(60*60*24*365); which will convert your seconds to years (not accurate, leap years will not be taken into consideration). Though your best option is checking out #James C's link and use his solutions. I just wanted to give some explanation.