Within my CodeIgniter app, I'm using a Jquery calendar pop-up that also captures time as set by the user, so the end result looks like: MM-DD-YYYY HH:MM, and I'm storing this in MySQL into a DateTime field that is: YYYY-MM-DD HH:MM:SS. What is the best (most efficient) way to push the date/time into MySQL so that it saves properly, and to pull is back out of MySQL and render it on the screen in the reverse format? Thanks!
Most efficient way is to use the ISO 8601 standard to pass date values between the client and server. Since the client and server talks in strings you'd be parsing the date to a string before sending it either way. The best format I prefer is the combined date and time in UTC:
2011-06-14T13:57Z
There are no spaces and it's clean. Then you'll have to parse it on the server side (should be relatively easy using PHP) and parse it on the client side.
For displaying purposes, I prefer to extend JavaScript's Date.prototype to include a format function that imitates PHP's date format.
Once you include the linked script from above you could do this on the server side -
var today = new Date();
alert(today.format('m-d-Y H:i')); //displays "06-14-2011 11:18"
Good luck!
I think you should use the strptime() function to parse the date received from the jQuery calendar your using and using mktime():
// Parse the time based on your jQuery calendar's format
$parts = strptime($calendar_value, '%m-%d-%Y %H:%M');
if ( ! empty($parts) )
{
// Create a Unix timestamp
$timestamp = mktime($parts['tm_hour'], $parts['tm_min'], 0, $parts['tm_mon'] + 1, $parts['tm_mday'], $parts['tm_year'] + 1900);
// Create a string representation of the Unix timestamp
$date = date(DATE_ISO8601, $timestamp);
}
You'll want to use $date to insert in your database. There is a function called "strtotime" which will attempts to parse dates that are in human-readable format but I doubt it's able to determine if the month or day comes first, especially if they're both lower than 12 which is why I chose to use "strptime" instead.
When you pull the data from MySQL, you can then simply use the date() and strtotime() function to populate the calendar:
echo date('m-d-Y h:i', strtotime($mysql_date));
Related
I am using a jquery datepicker with the following format mm/dd/yyyy but I need it to be yyyy-mm-dd for the sql db so I am using this.
$date = date("Y-m-d",strtotime($startdate));
with the following
$query = "INSERT INTO complete_table ( name, startdate) VALUES ('$_POST[name]', '$date')";
Unfortunately I register 1970-01-01 in the db regardless of what the input is.
Any idea on what I am doing wrong? Thank you so much for the help.
When you get 1970-01-01 back from date() it means that the timestamp is incorrect. That date is the UNIX epoch.
When it comes to Javascript (in this case a datepicker) you should always make sure that what you get is, in fact, what you expect it to be. Simply passing it through strtotime() will cause problems like the one you have here.
One good way to handle dates is using the datetime extension. It has a static method, DateTime::createFromFormat, that will let you parse any date using any format:
// The value from the datepicker
$dpValue = '12/30/2013';
// Parse a date using a user-defined format
$date = DateTime::createFromFormat('d/m/Y', $dpValue);
// If the above returns false then the date
// is not formatted correctly
if ($date === false) {
header('HTTP/1.0 400 Bad Request');
die('Invalid date from datepicker!')
}
// Using the parsed date we can create a
// new one with a formatting of out choosing
$forSQL = $date->format('Y-m-d');
// ...
put dateformat in jquery when initializing the datepicker like
$( "#id/.class" ).datepicker({dateFormat: 'yy-mm-dd'});
This format will insert on mysql db(done on client side), otherwise format it on the server side code.
$startdate is invalid and therefor the strtotime function returns anything below 86400 (most likely 0), which is the amount of seconds in one day. Please refer to the PHP documentation for the right formatting of $startdate here.
It is called the epoch. So startdate is zero.
I need to display and handle UTC dates in the following format:
2013-06-28T22:15:00Z
As this format is part of the ISO8601 standard I have no trouble creating DateTime objects from strings like the one above. However I can't find a clean way (meaning no string manipulations like substr and replace, etc.) to present my DateTime object in the desired format. I tried to tweak the server and php datetime settings, with little success. I always get:
$date->format(DateTime::ISO8601); // gives 2013-06-28T22:15:00+00:00
Is there any date format or configuration setting that will give me the desired string? Or I'll have to append the 'Z' manually to a custom time format?
No, there is no special constant for the desired format. I would use:
$date->format('Y-m-d\TH:i:s\Z');
But you will have to make sure that the times you are using are really UTC to avoid interpretation errors in your application.
If you are using Carbon then the method is:
echo $dt->toIso8601ZuluString();
// 2019-02-01T03:45:27Z
In PHP 8 the format character p was added:
$timestamp = new DateTimeImmutable('2013-06-28T22:15:00Z');
echo $timestamp->format('Y-m-d\TH:i:sp');
// 2013-06-28T22:15:00Z
In order to get the UTC date in the desired format, you can use something like this:
gmdate('Y-m-d\TH:i:s\Z', $date->format('U'));
To do this with the object-oriented style date object you need to first set the timezone to UTC, and then output the date:
function dateTo8601Zulu(\DateTimeInterface $date):string {
return (clone $date)
->setTimezone(new \DateTimeZone('UTC'))
->format('Y-m-d\TH:i:s\Z');
}
Edit: clone object before changing timezone.
Since PHP 7.2 DateTimeInterface::ATOM was introduced in favor of DateTimeInterface::ISO8601, although it still lives on for backward compatability reasons.
Usage
$dateTimeObject->format(DateTimeInterface::ATOM)
I want to input a timestamp in below format to the database.
yyyy-mm-dd hh:mm:ss
How can I get in above format?
When I use
$date = new Zend_Date();
it returns month dd, yyyy hh:mm:ss PM
I also use a JavaScript calender to insert a selected date and it returns in dd-mm-yyyy format
Now, I want to convert these both format into yyyy-mm-dd hh:mm:ss so can be inserted in database. Because date format not matching the database field format the date is not inserted and only filled with *00-00-00 00:00:00*
Thanks for answer
Not sure if this will help you, but try using:
// to show both date and time,
$date->get('YYYY-MM-dd HH:mm:ss');
// or, to show date only
$date->get('YYYY-MM-dd')
Technically, #stefgosselin gave the correct answer for Zend_Date, but Zend_Date is completely overkill for just getting the current time in a common format. Zend_Date is incredibly slow and cumbersome to use compared to PHP's native date related extensions. If you don't need translation or localisation in your Zend_Date output (and you apparently dont), stay away from it.
Use PHP's native date function for that, e.g.
echo date('Y-m-d H:i:s');
or DateTime procedural API
echo date_format(date_create(), 'Y-m-d H:i:s');
or DateTime Object API
$dateTime = new DateTime;
echo $dateTime->format('Y-m-d H:i:s');
Don't do the common mistake of using each and every component Zend Frameworks offers just because it offers it. There is absolutely no need to do that and in fact, if you can use a native PHP extension to achieve the same result with less or comparable effort, you are better off with the native solution.
Also, if you are going to save a date in your database, did you use any of the DateTime related columns in your database? Assuming you are using MySql, you could use a Timestamp column or an ISO8601 Date column.
This is how i did it:
abstract class App_Model_ModelAbstract extends Zend_Db_Table_Abstract
{
const DATE_FORMAT = 'yyyy-MM-dd';
public static function formatDate($date, $format = App_Model_ModelAbstract::DATE_FORMAT)
{
if (!$date instanceof Zend_Date && Zend_Date::isDate($date)) {
$date = new Zend_Date($date);
}
if ($date instanceof Zend_Date) {
return $date->get($format);
}
return $date;
}
}
this way you don't need to be concerned with whether or not its actually an instance of zend date, you can pass in a string or anything else that is a date.
a simple way to use Zend Date is to make specific function in its business objects that allows to parameter this function the date format. You can find a good example to this address http://www.pylejeune.fr/framework/utiliser-les-date-avec-zend_date/
this is i did it :
Zend_Date::now->toString('dd-MM-yyyy HH:mm:ss')
output from this format is "24-03-2012 13:02:01"
and you can modified your date format
I've always use $date->__toString('YYYY-MM-dd HH-mm-ss'); method in the past but today didn't work. I was getting the default output of 'Nov 1, 2013 12:19:23 PM'
So today I used $date->get('YYYY-MM-dd HH-mm-ss'); as mentioned above. Seems to have solved my problem.
You can find more information on this on output formats here: http://framework.zend.com/manual/1.12/en/zend.date.constants.html
I display the date or the time on my website a lot and I'm thinking about writing a function to parse a PostgreSQL timestamp.
The timestamp is in the format: Y-m-d H:i:s.u. E.g. 2011-04-08 23:00:56.544.
I'm thinking about something like this:
function parse_timestamp($timestamp, $format = 'd-m-Y')
{
// parse the timestamp
return $formatted_timestamp;
}
However I am wondering whether this can also be achieved without writing a parser for it myself (with the use of some PHP function).
function parse_timestamp($timestamp, $format = 'd-m-Y')
{
return date($format, strtotime($timestamp));
}
Don't forget to set timezone before, e.g.
date_default_timezone_set('UTC');
Or in your case, I guess 'Europe/Amsterdam'.
You can always get PHP timestamp of this format Y-m-d H:i:s.u using strtotime(). Then, using date() you can export time in your own format. Both functions depend of time zone set.
strtotime is perfectly capable of parsing that time string, then just reformat it with date:
echo date('d-m-Y', strtotime('2011-04-08 23:00:56.544')); // 08-04-2011
For those using DateTime class:
DateTime::createFromFormat('Y-m-d H:i:s.u', $yourTime);
If the database isn't giving you what you want, change it. PostgreSQL can also format dates and times.
select to_char(timestamp '2011-03-04 07:04:00', 'DD-MM-YYYY');
04-03-2011
But that's risky in international contexts, like the web. Different locales expect different ordering of elements. (Does "04-03" mean 03-April or 04-March?) This expression is easier to understand, and it uses locale-specific abbreviations for the months.
select to_char(timestamp '2011-03-04 07:04:00', 'DD-Mon-YYYY');
04-Mar-2011
Take a look at the strptime function - it can parse many time formats.
To change 2009-12-09 13:32:15 to 09/12/2009
here:
echo date("d/m/Y", strtotime('2009-12-09 13:32:15'))
You can use strtotime to get the timestamp of the first date, and date to convert it to a string using the format you want.
$timestamp = strtotime('2009-12-09 13:32:15');
echo date('d/m/Y', $timestamp);
And you'll get :
09/12/2009
[edit 2012-05-19] Note that strtotime() suffers a couple of possibly important limitations:
The format of the date must be YYYY-MM-DD; it might work in some other cases, but not always !
Also, working with UNIX Timestamps, as done with date() and strtotime() means you'll only be able to work with dates between 1970 and 2038 (possibly a wider range, depending on your system -- but not and illimited one anyway)
Working with the DateTime class is often a far better alternative:
You can use either DateTime::__construct() or DateTime::createFromFormat() to create a DateTime object -- the second one is only available with PHP >= 5.3, but allows you to specify the date's format, which can prove useful,
And you can use the DateTime::format() method to convert that object to any date format you might want to work with.
Using the date() method.
print date("d/m/Y", strtotime("2009-12-09 13:32:15"));
$long_date = '2009-12-09 13:32:15';
$epoch_date = strtotime($long_date);
$short_date = date('m/d/Y', $epoch_date);
The above is not the shortest way of doing it, but having the long date as an epoch timestamp ensures that you can reuse the original long date to get other date format outputs, like if you wanted to go back and have just the time somewhere else.