I need to populate a table with a row for each day over the past 5 years?
E.g
1. 04/11/2012
2. 03/11/2012
3. 02/11/2012
4. 01/11/2012
5. 31/10/2012
6. etc
This will just be a php script that I run once and inserts all of the records.
The reason for doing this is for performance testing my other code with enough data in the database.
This will give you the date-format you want:
date('d/m/Y', strtotime($i." days"));
$i is the number of days past/ahead of this date. So if $i is -1 that means one day ago. I guess you could make a loop of it, decreasing/increasing $i.
Like this?
$date = "2007-01-01";
while ($date != "2013-01-01") {
//do your query... not going to do any escaping for brevity
mysql_query("INSERT INTO TABLENAME (SomeDateColumn) VALUES ('$date')");
$date = date("Y-m-d", strtotime($date) + 86400);
}
$now = new DateTime(); // Now
$date = $new DateTime();
$date->sub(new DateInterval('P5Y')); // 5 years ago
while ($date->diff($now))->invert != 1)
{
// insert $date into database
// I'll leave that to you, as you were looking for a way to loop over dates
$date->add(new DateInterval("P1D"));
}
Related
I'm working on a tick based space game http://ricarion.com/ but the ticks only run between certain hours.
08:00-16:30 - run every 30 minutes via a cron job. In the nav bar at the top I want to add "Next Tick: 08:30 06/02/20" for example.
So I was thinking of creating an array:
$tick_times[] = array();
$tick_times[] = 08:00;
$tick_times[] = 08:30;
$tick_times[] = 09:00;
...
$tick_times[] = 16:30;
And then this is where I get stuck, how do I check the existing time, and then compare that against the array selecting the next future time? i.e. It's now 08:34, so the return should be 09:00?
Did you need an array or just want to calculate the next 30-minute interval?
If so this may be similar to:
Round minute down to nearest quarter hour
You do modulo division of 1800 seconds on the current time and add the answer (time remainder of time to the next interval) to the then-current time to get the next event.
<?php
$current_date = date('d-M-Y g:i:s A');
echo $current_date."\n";
$current_time = strtotime($current_date);
$frac = 1800;
$r = $current_time % $frac;
$new_time = $current_time + ($frac-$r);
$new_date = date('d-M-Y g:i:s A', $new_time);
echo $new_date."\n";
http://codepad.org/xs9lMCRQ
Get the now time format it and compare it. In your case you maybe format your $tick_time to the same format like current time.
$date = new DateTime('now');
$date = $date->format('Y-m-d H:i:s');
foreach ($tick_times as $tick_time) {
$date_added = new DateTime($tick_time);
if (strtotime($date_added) == strtotime($date)) {
//do your stuff here
}
}
This question already has answers here:
Adding days to $Date in PHP
(12 answers)
Closed 5 years ago.
I have a table in MySQL with a date field (called NDate) which contains standard date values ("2017-04-17","2017-04-18", etc.).
Through PHP webpage, I am trying to take the system date (say today is 2017-04-17), and then pull all rows from the above table where NDate="2017-04-17". No issues till here.
I have a requirement to increment the day (starting today and going on for next 10 days - i.e. 2017-04-17 to 2017-04-26), and for each day report entries under a different heading like "Entries for 2017-04-17" which will list all rows having NDate 2017-04-17, "Entries for 2017-04-18" which will list all rows having NDate 2017-04-18.
I was trying to use a for loop with PHP date_modify function to increment the days one by one, but it is not showing any results.
Here are the selected pieces of code:
date_default_timezone_set('US/Eastern');
$datev = date("Y-m-d");
for ($x = 0; $x <= 10; $x++)
{
$datev=date_modify($date,"+$x days");
echo "before date format<br>"; // echo statement 1
echo "date is: $datev <br>"; // echo statement 2
$sql = "SELECT * FROM tablename where Ndate='$datev'";
echo "before result<br>"; // echo statement 3
...
...
...
}
Output on webpage shows only statement 1. But echo stats 2 and 3 are not printed.
You can increment days using strtotime function as a parameter to date function.
For 10 days, you can use for loop, to build an array of days. Then iterate over it, to execute queries you need.
$today = date('Y-m-d');
$dates=array($today);
for($i=1;$i<10;$i++) {
$NewDate=date('Y-m-d', strtotime("+".$i." days"));
$dates[]=$NewDate;
}
foreach($dates as $dt) {
// sql stuff here
echo "date is: $dt <br>";
$sql = "SELECT * FROM tablename where Ndate='$dt'";
echo "before result<br>";
// .....
}
This code should work for your case. If any problems, just let me know.
Try this:
$start = strtotime(date('Y-m-d'));
$end = strtotime(date('Y-m-d', strtotime('+10 days')));
while($start <= $end)
{
$date = date('Y-m-d', $start);
//use $date to do stuff
//SELECT * FROM tablename where Ndate='$date'
$start = strtotime("+1 day", $start);
}
I want to loop a date so that every time date is increment by previous date. my code is here. plz reply anyone, thanks in advance
$today = date('Y-m-d');
for($i=1; $i<=4; $i++){
$repeat = strtotime("+2 day",strtotime($today));
echo $rdate = date('Y-m-d',$repeat);
}
I want result as if today is 2016-04-04 than, 2016-04-06, 2016-04-08, 2016-04-10, 2016-04-12.
actually i want to make a reminder date where user enter reminder. lets a user want to add reminder today and want repeat it 5 time after 2days, 3days or what ever he wants, in next comming day. than how i repeat date with for loop.
Try this:
<?php
$today = date('Y-m-d');
for($i=1; $i<=4; $i++)
{
$repeat = strtotime("+2 day",strtotime($today));
$today = date('Y-m-d',$repeat);
echo $today;
}
Output:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
The easiest way is what answer
aslawin
The below example is to go through the date
$begin = new DateTime($check_in);
$end = new DateTime($check_out);
$step = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($begin, $step, $end);
foreach ($period as $dt)
{
<sample code here>
}
You can try this:
$today = date('Y-m-d');
for($i=1; $i<=8; $i++){
if($i%2 == 0){
$repeat = strtotime("+$i day",strtotime($today));
echo $rdate = date('Y-m-d',$repeat);
}
}
Result:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
In this example, you can use $i%2 == 0 with limit <= 8
Use a for loop with base 2, then directly output your dates:
for( $i=2; $i<9; $i=$i+2 )
{
echo date('Y-m-d', strtotime( "+ $i days" )) . PHP_EOL;
}
Result:
2016-04-06
2016-04-08
2016-04-10
2016-04-12
actually i want to make a reminder date where user enter reminder.
lets a user want to add reminder today and want repeat it 5 time after
2days, 3days or what ever he wants, in next comming day. than how i
repeat date with for loop.
I'll help with the above. First of all I will just say I have a huge personal preference towards the DateTime object over simply using date it's more flexible and a hell of a lot more readable in my opinion, so when working with dates I would always suggest using that over date()
So here is some Code:
$date = new DateTime(); // Pretend this is what the User entered. We got it via $_POST or something.
$interval = 2; // Repeat x times at y day intervals. (Not including the initial)
$repeatAmount = 2; // Repeat the reminder x times
for ($i = 0; $i <= $repeatAmount; ++$i) {
echo $date->format('d/m/Y');
$date->modify('+'. $interval .' day');
}
$date = new DateTime()Imagine this is the date the user entered, this is our starting point, our first reminder will at this time.
$interval and $repeatAmount are the interval in days, i.e. I want this to every 2 days and the amount of times you want it to repeat, in our example 2.
for ($i = 0; $i <= $repeatAmount; ++$i) { We want to loop as many times as the user says they want to repeat. Little note ++$i tends to be a very minor performance boost over $i++ in some scenarios, so it is usually better to default to that unless you specifically need to use $i++
echo $date->format('d/m/Y'); Simply print out the date, i'll let you handle the reminder logic.
$date->modify('+' . $interval . ' day'); Increment the dateTime object by the interval that the user has asked for, in our case increment by 2 days.
Any questions let me know.
I am currently attempting to get a list of dates from a current date using the following format so that I can process it and stick it in my database
Saturday/02-05-2015
So far, i've managed to get the system to output the date correctly, but can not get it to increment in single day values.
My current code to attempt to increment this is the following
$tempStartDateN = ("$splode[0]/$splode[1]/$splode[2]/$splode[3]");
echo $tempStartDateN;
$tempStartDateN = date('l/d/m/Y', strtotime($tempStartDateN . ' + 1 day'));
echo $tempStartDateN;
I am currently using explode to process the data after the increment, which works fine, but can not get the date itself to increment as long as the day name is included.
Currently, the time is got using this code, which is processed afterwords using explode
$OldDateArray = date("Y/m/d/l");
So to keep a long question short, what is the best way to increment a date that requires the day name, day, month then year?
EDIT:
Heres my current code, managed to get this far thanks to SamV
$date = date("l/d/m/Y");
echo $date;
echo ('</br>');
list($weekdayName, $dateString) = explode("/", $date, 2);
$dateObj = new \DateTime($dateString);
for($i=0; $i<=5; $i++){
$dateObj->add(new \DateInterval("P1D")); // P1D stands for "Period 1 Day"
echo $dateObj->format("l/d/m/Y"); // Sunday/03/05/2015
echo ('</br>');
}
What this does however is:
Friday/01/05/2015
Tuesday/06/01/2015
Wednesday/07/01/2015
Thursday/08/01/2015
Friday/09/01/2015
Saturday/10/01/2015
Sunday/11/01/2015
this means that date and month are swapping around, what is causing this?
You don't need to parse the week day name to add days onto a date.
$date = "Saturday/02-05-2015";
list($weekdayName, $dateString) = explode("/", $date, 2); // Parse "02-05-2015"
$dateObj = new \DateTime($dateString);
$dateObj->add(new \DateInterval("P1D")); // P1D stands for "Period 1 Day"
echo $dateObj->format("l/d/m/Y"); // Sunday/03/05/2015
I used the DateTime class, here is the documentation.
I wrote out what you are trying to do yourself, not sure what is causing your issue. This code works though.
$date = "Friday/01-05-2015";
list($weekdayName, $dateString) = explode("/", $date, 2); // Parse "01-05-2015"
$dateObj = new \DateTime($dateString);
for($i = 0; $i < 5; $i++) {
$dateObj->add(new \DateInterval("P1D")); // P1D stands for "Period 1 Day"
echo $dateObj->format("l/d/m/Y") . '<br>';
}
Outputs:
Saturday/02/05/2015
Sunday/03/05/2015
Monday/04/05/2015
Tuesday/05/05/2015
Wednesday/06/05/2015
If strtotime is able to parse a date it returns the timestamp. Why not add to it the number of seconds in a day? Smth. like $timestamp += 24 * 3600;
P.S. As far as I can understand, strtotime may accept timestamp as second argument (http://us2.php.net/manual/en/function.strtotime.php) smth. like $timestamp = strtotime('+1 day', $timestamp);
Good Day,
I am trying to create a recurring date system that has the following:
nth day of nth month (2nd day of every 3rd month)
$this_months_friday = strtotime('+3 days +4 months');
the output of that will always be current day + 3 days of the 4th month.
how do I get it to display the nth day of the nth month?
since i also tried
$this_months_friday = strtotime('every 3 days +4 months');
and it did not return any result. Should i stick with strtotime on this one or move to DateTime function of php. though i wont still be able to formulate the proper argument for that kind of date sequence.
Any help would be greatly appreciated.
Thank You.
Probably better off using DateTime with a couple intervals:
$d = new DateTime();
$d->add(new DateInterVal('P' . $days . 'D'))->add('new DateInterVal('P' . $months . 'M'));
not sure what youre two example intervals are wanting.
You want an internval to start in 4 months, which then repeats every 3 days?
That'd be something more like
$d = new DateTime();
$d->add(new DateInterval('P4M')); // jump ahead 4 months immediately
$day3 = new DateInterval('P3D');
for ($i = 0; $i < 100; $i++) {
$d->add($day3); // jump ahead 3 days
... do something with this new date
}
for a basic recurring event, +4 months + 3 days, you'd simply have one interval:
$interval = new DateInteval('P4M3D'); // +4 months +3 days
$date = new DateTime();
while($some_condition) {
$date->add($interval);
do_something();
}
You can do this by saving the values in variables like that :
$day=3;
$month=4;
echo date("d-m-y",strtotime('+'.$day .'days' .'+'.$month.'months'));
Explanation:
7(july)+4 months = 11 month(November)
8 july+ 3 days = 11 july
Output:
11-11-13
NOTE: just for the example I have put the values hard coded, You can make them dynamic.