I'm creating a basic game for a final project; the entire site is built and ready to go, but I can't get one functionality working.
The function is that over time the user's pet's stats will lower.
The longer you are away from your pet the lower their stats get until they "die". So far I'm trying to calculate the time difference between the logins by using this code:
$mod = mysqli_query($con,"select * from users where user_name='$user_name'");
while($row = mysqli_fetch_array($mod)){
$last_login = $row['last_login'];
$stored_login = $row['stored_login'];
$time_diff = ($last_login-$stored_login);
echo $time_diff;
$time_mod = mysqli_query($con,"update users set time_mod='$time_diff' where user_name='$user_name'");
}
Both the last_login and stored_login are TIMESTAMP variables, and I've been told I should be able to subtract them easily to get a solution.
However, when I echo the equation all it returns is 0. I've tried using DATEDIFF but MySQL gives me an error saying I can't use it. I only need the number of days that have passed - Is there any way to make this happen?
If it helps, you can access the beta here (either create an account or you can log on with):
http://www.eurogabby.com/MyPetMonster/login.php
username: a
password: a
Maybe you can try. Convert to time the row.
$last_login = strtotime($row['last_login']);
$stored_login = strtotime($row['stored_login']);
$time_diff = ($last_login - $stored_login);
echo $time_diff;
you just need to convert the time using strtotime like this
$time_diff = (strtotime($last_login)-strtotime($stored_login));
echo $time_diff;
if this not works then first convert them to strtotime and store in different variables then subtract them like
$last = strtotime($last_login);
$stored = strtotime($stored_login);
$time_diff = ($last-$stored);
echo $time_diff;
Now you can convert back to time using date function like this
$newtime = date("m/d/Y",$time_diff);
echo $newtime = month/day/year
remember m/d/y will be your choice of date format in which you want to convert.
Related
I have been trying to figure this out for a week now. My wife has started a new taxi-company and she asked me to code a simple webpage for here where she could press a button to save a timestamp, then the press is again when she gets off work, it then creates a second timestamp
I have an MYSQL database with rows containing the start time and stop time. I have managed to use the diff function to see how much time it is between the two timestamps but now comes the tricky part.
Since it's different payments at different times of the day I need to divide the time at a shortened time.
Up to 19:00 she works "daytime" and after that, she works "nighttime" until 06:00 the other day, then there is "weekend daytime" and "weekend nighttime" as well.
So if she creates a timestamp whit the date and time: 2018-08-08 06:30 and then another timestamp at 2018-08-08 21:00, then I need a script that puts these data in ex "$daytimehours = 12" "$daytimeminutes = 30" and "$nighttimehours = 3" "$nighttimeminutes = 0"
I have managed to create a script that almost works, but it is several pages long, and it contains one if-statement for each different scenario daytime-nighttime, nighttime-daytime etc.
So do anyone has a good idea on how to solve this? or maybe just point me in the right direction. I would be happy to pay some money to get this to work.
My solution is
<?php
date_default_timezone_set('Asia/Almaty');
$endDate = '2018-08-08 21:00';
$startDate = '2018-08-08 06:30';
$nightmare = date('Y-m-d 19:00');
$startDay = date('Y-m-d 06:00');
$diffMorning = strtotime($nightmare) - strtotime($startDate);
$diffNight = strtotime($endDate) - strtotime($nightmare);
echo gmdate('H:i', $diffMorning) . "\n"; // this is the difference from start day till 19:00
echo gmdate('H:i', $diffNight); // this is the difference on nightmare
$total = $diffMorning + $diffNight;
echo intval($total/3600) . " hours \n";
echo $total%3600/60 . " minutes \n";
echo $total%3600%60 . ' seconds';
You can check via online compiler
given two dates stated as:
$endDate = '2018-08-08 21:00';
$startDate = '2018-08-08 06:30';
you can use the PHP Date extension to achieve the difference like this:
$start = date_create($startDate);
$end = date_create($endDate);
$boundnight = clone($end);
$boundnight->setTime(19,00);
$total_duration = date_diff($end,$start);//total duration from start to end
$day_duration = date_diff($boundnight,$start);//daytime duration
$night_duration = date_diff($end,$boundnight);// nighttime duration
you can use the format method to print a human readable string this way:
$total_duration=$total_duration->format('%H:%I');
$day_duration=$day_duration->format('%H:%I');
$night_duration=$night_duration->format('%H:%I');
At this step there is nothing left but you say you want to convert each duration in minutes.So let's build a function :
function toMinute($duration){
return (count($x=explode(':',$duration))==2?($x[0]*60+$x[1]):false);
}
Then you can use it this way:
$total_duration = toMinute($total_duration);
$day_duration = toMinute($day_duration);
$night_duration = toMinute($night_duration);
The output of
var_dump($total_duration,$day_duration,$night_duration) at this step is:
int(870)
int(750)
int(120)
I want to check if 30 min passed after created time in database. created is a time column having time stamp in this format 1374766406
I have tried to check with date('m-d-y H:i, $created) but than of course it is giving human readable output so don't know how to perform check if current time is not reached to 30min of created time.
Something like if(created > 30){}
Try this:
$created = // get value of column by mysql and save it here.
if ($created >= strtotime("-30 minutes")) {
// its over 30 minutes old
}
The better approach is to use DateTime for (PHP 5 >= 5.3.0)
$datenow = new DateTime();
$datenow->getTimestamp();
$datedb = new DateTime();
$datedb->setTimestamp(1374766406);
$interval = $datenow->diff($datedb);
$minutes = $interval->format('%i');
$minutes will give you the difference in minutes, check here for more
http://in3.php.net/manual/en/datetime.diff.php
Here is the working code
http://phpfiddle.org/main/code/jxv-eyg
You need to use strtotime(); to convert the date in human form back to a timestamp, then you can compare.
EDIT: Maybe I misread.
So something like;
if(($epoch_from_db - time()) >= 1800){
//do something
}
How can I compute time difference in PHP?
example: 2:00 and 3:30.
I want to convert the time to seconds then subtract them then convert it back to hours and minutes to know the difference. Is there an easier way to get the difference?
Look at the PHP DateTime object.
$dateA = new DateTime('2:00');
$dateB = new DateTime('3:00');
$difference = $dateA->diff($dateB);
(assuming you have >= PHP 5.3)
You can also do it the procedural way...
$dateA = strtotime('2:00');
$dateB = strtotime('3:00');
$difference = $dateB - $dateA;
See it on CodePad.org.
You can get the hour offset like so...
$hours = $difference / 3600;
If you are dealing with times that fall between a 24 hour period (0:00 - 23:59), you could also do...
$hours = (int) date('g', $difference);
Though that is probably too inflexible to be worth implementing.
Check this link ...
http://www.onlineconversion.com/days_between_advanced.htm
I used this to calculate the difference between server time and the users local time. Grab the hour difference and drop that in a form when the user is registering. I then use it to update the time on the site for the user when they do stuff online.
Once I got it working, I switched this line ...
if (form.date1.value == "")
form.date1.value = s;
to ...
form.date1.value = "<?PHP echo date("m/d/Y H:i:s", time()) ?>";
Now I can compare the user time and the server time! You can grab the seconds and mins as well.
I have a MySQL DB with StartDates in the format of yyyy-mm-dd and starttimes in the format of HH:MM using a 24 hour clock. What would be the easiest way to compare the difference of two days in PHP? Would using a datetime object could I set it to be with the information given and just give it to zeros for the seconds? I need to get the amount of time between both dates down to the minute. I was putting the startdate (Just the day since its always within the same month for my application) and time together concatenated together and then pulling out what I need like below, but I haven't been able to get it straight yet. Thanks for the look!
$tempvar1 = $times[$i][$j];
$tempvar2 = $times[$i][$j+1];
$day1 = $tempvar1[0].$tempvar1[1];
$day2 = $tempvar2[0].$tempvar2[1];
$hours1 = $tempvar1[2].$tempvar1[3];
$hours2 = $tempvar2[2].$tempvar2[3];
$minutes1 = $tempvar1[5].$tempvar1[6];
$minutes2 = $tempvar2[5].$tempvar2[6];
$numdays = ($day2-$day1) - 1;
$time1 = ($hours1*60)+$minutes1;
$time2 = ($hours2*60)+$minutes2;
MySQL has plenty of date/time functions:
SELECT TIMEDIFF(endtime, starttime), DATEDIFF(endtime, starttime)
FROM ...
doc links for timediff and datediff
That'll you get strings in the format of 'hh:mm:ss.ssss' for timediff, and a straight-up integer representing the days between the two dates, respectively.
I have a function which checks my database to see if a date exists, if it does exist, i want to display the next date which isnt in the database.
Is this possible?
My function returns 1 if there is a date in the database and 0 if there isnt, im using codeigniter, but not using any built in functions.
Its basically an availability checker, it allows us to input many different dates in the database, so calling my function i use
$availcheck = $ci->availability->check_availability_by_date(date('d/m/Y'));
The i use a if statement to check if the first time it runs it returns a value, this is how i have it
if($availcheck > 0){
// loop through the next dates and run the function again to see if it returns 0
} else {
echo 'available now';
}
I guess i would add 1 to the current date, check that one, then add another 1 and check that and so on.
Im just not sure how.
Cheers,
if i understand you correct , your problem is adding the day ?
if so i would suggest using the epoch or unix time
so convert the date to unix time using mktime than just add 1 day in seconds (24*60*60)
and then convert back to d/m/y format.
you can use the date function.
$date = time(); // get current timestamp
while ($availcheck) // while date IS found in database
{
$availcheck = $ci->availability->check_availability_by_date(date('d/m/Y',$date));
$date = $date + (24*60*60); // add one day
}
$date = $date - (24*60*60); // reduce one day
echo date('d/m/Y',$date); // prints the first date that is not in the DB
This SQL code could work for me.
$today = date("Y-m-d"); //today
$sql = "SELECT date FROM calendar WHERE date>'{$today}' AND date<='2100-12-31' AND date='0000-00-00' LIMIT 1";
Since you can't determine the ending date, 2100 could be for testing.