I want to convert this date ( 02-12-2010) mm-dd-yyyy to time format
ie
to 02-12-2010 0 hours 0 minutes and 0 seconds
i have user time and date functions but when i refresh the page its value is changing as per the time.
i need that to be fixed
also i want to convert this date (01-24-2009) to time format.
please help me
Thanks
Use the strtotime function to convert an existing date/time string into a timestamp for the date function.
$new_date = date('m-d-Y h:i:s', strtotime('02-12-2010'));
The reason that your date keeps updating with the current time is that the date() function uses the current system's timestamp by default when no second parameter is provided.
Use the date_parse_from_format () API function to transform your given format to an array. E.g.
$date = "02-12-2010";
$dateArr = date_parse_from_format("d-m-Y", $date);
Output it with something like:
$output = $dateArr['day'] . '-' . $dateArr['month'] '-' . $dateArr['year'];
You can use the strtotime function eg:
$mydate = date('m-d-Y h:i:s', strtotime('02-12-2010'));
Related
I have SQLite DB one table contains datetime field
with datatype "timestamp" REAL value is 18696.0
attach image for table structure
So, I want this 18696.0 value to be converted into MySQL Y-m-d format and result should be 2021-03-10
I have didn't found any solution online. any help would be appreciated.
SQLite timestamp converted into MySQL timestamp.
EDIT: Thankyou for updating your question with the correct number and what date it should represent.
You can achieve what you need with a function that adds the days onto the Unix Epoch date:
function realDateToYmd($real, $outputFormat='Y-m-d')
{
$date = new DateTime('1970-01-01');
$date->modify('+' . intval($real) . ' days');
return $date->format($outputFormat);
}
echo realDateToYmd('18696.0');
// returns 2021-03-10
SQLite dates stored in REAL data type stores dates as a Julian Day.
From https://www.sqlite.org/datatype3.html
REAL as Julian day numbers, the number of days since noon in Greenwich on November 24, 4714 B.C. according to the proleptic Gregorian calendar.
PHP has a jdtogregorian function, in which one comment has a handy function to convert to ISO8601 dates:
function JDtoISO8601($JD) {
if ($JD <= 1721425) $JD += 365;
list($month, $day, $year) = explode('/', jdtogregorian($JD));
return sprintf('%+05d-%02d-%02d', $year, $month, $day);
}
echo JDtoISO8601('17889.0');
// Results in -4664-11-16
The results don't exactly look right, is it definitely 17889.0 in SQLite?
If this float number 18696.0 represents the number of days since 1970-01-01 then the date can also be calculated like this:
$days = 18696.0;
$dt = date_create('#'.((int)($days * 86400)));
$mysqlDate = $dt->format('Y-m-d'); //"2021-03-10"
background information
Or simply with gmdate:
$mySqlDate = gmdate('Y-m-d',$days*86400);
The days are simply converted into seconds to get a valid timestamp for gmdate.
Try this:
<?php
echo date('Y-m-d H:i:s', 17889);
?>
Output:
1970-01-01 04:58:09
For eg I have ISO date time as : 2020-03-03T11:07:41.1708478Z
Converting using strtotime function
$dateTime = date("Y-m-d H:i:s.u",strtotime('2020-03-03T11:07:41.1708478Z'));
Result : 2020-03-03 11:07:41.000000
In above result you can see milliseconds are gone.
Use DateTime because strtotime and date only uses full seconds.
$date = new DateTime('2020-03-03T11:07:41.1708478Z');
echo $date->format("Y-m-d H:i:s.u"); // 2020-03-03 11:07:41.170847
https://3v4l.org/DXVj8
But since this I assume this input format is rater fixed and wont change I could recommend you to use a simple str_replace.
echo str_replace(["T","Z"], [" ",""], '2020-03-03T11:07:41.1708478Z'); // 2020-03-03 11:07:41.170847
I want to know how to pick the time format from datetime in yii2.
So if there is code like this
$model->date = date('Y-m-d H:i:s', strtotime('+5 hours'));
and the result would be 2018-05-08 23:36:21
how can I extract the time only? so the result is only 23:36:21
I already tried using code below
date("H:i:s", strtotime('-30 minutes'));
but I only got like 00:30:00
Is there something wrong with the code?
Any help would be appreciated :)
if you have the string 2018-05-08 23:36:21 and want to extract time, you can follow the following methods
Using php:date() function
Remove -30m from the time when you format the date
echo date('H:i:s',strtotime('2018-05-08 23:36:21 -30 minutes'));
Using DateTime Object
The method sub() can subtract from the time and output the remaining time using $dateTimeObj->format().
$date1=new \DateTime('2018-05-08 23:36:21');
$date1->sub(new \DateInterval('PT30M'));
echo $date1->format('H:i:s');
I am using DateTime function of php. I get a date from a calendar in format d-m-Y and pass it via ajax to my function. I am getting the date right till this step.
When I try to store the date in unix format using:
$ai_ff_date=DateTime::CreateFromFormat('d-m-Y', $data['date']);
$final_date=$ai_ff_date->format('U');
The date stored is wrong. Suppose the date I passed via ajax is 26-12-2016 then in database 27-12-2016 is stored. Why its counting one more day then the input.
use this code :
$date = date('Y-m-d H:i:s', strtotime('-1 day', $stop_date));
$ai_ff_date=DateTime::CreateFromFormat('d-m-Y',$date);
$final_date=$ai_ff_date->format('U');
and please check the variable (code not tested)
You might want to convert the Date-Format to "Y-m-d" First and then call-in the DateTime() Constructor. However, since what you are trying to do is just get the TimeStamp you might also do that directly without using DateTime. The Snippet below shows what is meant here:
<?php
$data = ['date'=>"13-12-2016"]; //<== JUST AN EXAMPLE FOR TESTING!!!
// SIMPLY CONVERT THE DATE TO Y-m-d FIRST.
$dateYMD = date("Y-m-d", strtotime($data['date']));
// THEN USE DateTime CONSTRUCTOR TO CREATE A NEW DateTime INSTANCE
// AND THEN RUN THE FORMAT YOU WISH::
$final_date = (new DateTime($dateYMD))->format('U');
var_dump($final_date); //<== YIELDS: string '1481583600' (length=10)
var_dump(date("Y-m-d", $final_date)); //<== YIELDS: string '2016-12-13' (length=10)
I have a timestamp of "1377592503467" stored in a variable and when I pass it to php's date() function, it returns a value of "1935-05-26 03:04:11". Hopefully I am missing something obvious; below is my code. The timestamp represents 2013-08-27 15:57:45 but that's not what is being returned.
$date = "1377592503467";
$formattedDate = date("Y-m-d h:i:s", $date);
Thank you.
Looks like that timestamp is in milliseconds, not seconds as PHP uses them.
Do date(..., $date / 1000).