In PHP: <?php define(‘TIMEZONE’, ‘Europe/Paris’); date_default_timezone_set(TIMEZONE); For MySQL: <?php $now = new DateTime(); $mins = $now->getOffset() / 60; $sgn = ($mins < 0 ? -1 : 1); $mins = abs($mins); $hrs = floor($mins / 60); $mins -= $hrs * 60; $offset = sprintf(‘%+d:%02d’, $hrs*$sgn, $mins); //Your DB Connection – sample $db = new PDO(‘mysql:host=localhost;dbname=test’, ‘dbuser’, … Read more
/** * Returns the week number for this date. dowOffset is the day of week the week * “starts” on for your locale – it can be from 0 to 6. If dowOffset is 1 (Monday), * the week returned is the ISO 8601 week number. * @param int dowOffset * @return int */ Date.prototype.getWeek … Read more
You likely have one or more docs with a created_at value that’s not a BSON Date and you’ll need to fix that by converting those values to Date or removing them. You can find those docs with a $not query that uses the $type operator like: db.snippets.find({created_at: {$not: {$type: 9}}}) If the created_at values are … Read more
To check if a date is valid you can parse the components of the date, create a Date object from it, and check if the components in the data is the same as the parsed components. If you create a Date object from compnents that are out of range, the values will flow over to … Read more
The JDBC specification does not define any details with regards to time zone. Nonetheless, most of us know the pains of having to deal with JDBC time zone discrepencies; just look at all the StackOverflow questions! Ultimately, the handling of time zone for date/time database types boils down to the database server, the JDBC driver … Read more
Those methods have been deprecated. Instead, use the Calendar class. import java.text.DateFormat; import java.text.ParseException; import java.text.SimpleDateFormat; import java.util.Calendar; public final class DateParseDemo { public static void main(String[] args){ final DateFormat df = new SimpleDateFormat(“MM/dd/yyyy”); final Calendar c = Calendar.getInstance(); try { c.setTime(df.parse(“04/12/2011”)); System.out.println(“Year = ” + c.get(Calendar.YEAR)); System.out.println(“Month = ” + (c.get(Calendar.MONTH))); System.out.println(“Day = ” … Read more
why not use date() just like below,try this $t = strtotime(‘20130409163705’); echo date(‘d/m/y H:i:s’,$t); and will be output 09/04/13 16:37:05
With lubridate library(lubridate) ymd(“2010/03/17”) – years(2)
Some thing like this could be used: select dateofbirth from customer Where DateofBirth BETWEEN date(‘1004-01-01’) AND date(‘1980-12-31’); select dateofbirth from customer where date(dateofbirth)>date(‘1980-12-01’); select * from customer where date(dateofbirth) < date(‘now’,’-30 years’); If you are using Sqlite V3.7.12 or greater Dont use date(dateofbirth) just use dateofbirth. So your query would look like this: select * … Read more
You’ll probably find you’re setting the date to Feb 31, 2009 (if today is Jan 31) and Javascript automagically rolls that into the early part of March. Check the day of the month, I’d expect it to be 1, 2 or 3. If it’s not the same as before you added a month, roll back … Read more