You could use the TIMEDIFF() and the TIME_TO_SEC() functions as follows: SELECT TIME_TO_SEC(TIMEDIFF(‘2010-08-20 12:01:00’, ‘2010-08-20 12:00:00’)) diff; +——+ | diff | +——+ | 60 | +——+ 1 row in set (0.00 sec) You could also use the UNIX_TIMESTAMP() function as @Amber suggested in an other answer: SELECT UNIX_TIMESTAMP(‘2010-08-20 12:01:00’) – UNIX_TIMESTAMP(‘2010-08-20 12:00:00’) diff; +——+ | … Read more
Here’s an example of how to convert DATETIME to UNIX timestamp: SELECT UNIX_TIMESTAMP(STR_TO_DATE(‘Apr 15 2012 12:00AM’, ‘%M %d %Y %h:%i%p’)) Here’s an example of how to change date format: SELECT FROM_UNIXTIME(UNIX_TIMESTAMP(STR_TO_DATE(‘Apr 15 2012 12:00AM’, ‘%M %d %Y %h:%i%p’)),’%m-%d-%Y %h:%i:%p’) Documentation: UNIX_TIMESTAMP, FROM_UNIXTIME
If you are using the JPA annotations, you can use @PrePersist and @PreUpdate event hooks do this: @Entity @Table(name = “entities”) public class Entity { … private Date created; private Date updated; @PrePersist protected void onCreate() { created = new Date(); } @PreUpdate protected void onUpdate() { updated = new Date(); } } or you … Read more
You can alternatively use the DateTime class, which internally represents the time components independently. Thus it is not susceptible to the 2038 limitation (unless you use ::getTimestamp).
I’ve got it figured out for my purposes. I’ll summarize what I learned (sorry, these notes are verbose; they’re as much for my future referral as anything else). Contrary to what I said in one of my previous comments, DATETIME and TIMESTAMP fields do behave differently. TIMESTAMP fields (as the docs indicate) take whatever you … Read more
Using the Pandas type Timedelta available since v0.15.0 you also can do: In[1]: import pandas as pd In[2]: df = pd.DataFrame([ pd.Timestamp(‘20150111’), pd.Timestamp(‘20150301’) ], columns=[‘date’]) In[3]: df[‘today’] = pd.Timestamp(‘20150315’) In[4]: df Out[4]: date today 0 2015-01-11 2015-03-15 1 2015-03-01 2015-03-15 In[5]: (df[‘today’] – df[‘date’]).dt.days Out[5]: 0 63 1 14 dtype: int64
You can readily do this this with the input format U.u. $now = DateTime::createFromFormat(‘U.u’, microtime(true)); echo $now->format(“m-d-Y H:i:s.u”); This produces the following output: 04-13-2015 05:56:22.082300 From the PHP manual page for date formats: U = Seconds since the Unix Epoch u = Microseconds http://php.net/manual/en/function.date.php Thanks goes to giggsey for pointing out a flaw in my … Read more
datetime.datetime.fromtimestamp() is correct, except you are probably having timestamp in miliseconds (like in JavaScript), but fromtimestamp() expects Unix timestamp, in seconds. Do it like that: >>> import datetime >>> your_timestamp = 1331856000000 >>> date = datetime.datetime.fromtimestamp(your_timestamp / 1e3) and the result is: >>> date datetime.datetime(2012, 3, 16, 1, 0) Does it answer your question? EDIT: … Read more
Update June 2014: David Fraser mentions in the comments a solution also detailed in “Change timestamps while rebasing git branch“, using the option –committer-date-is-author-date (introduced initially in Jan. 2009 in commit 3f01ad6 Note that the –committer-date-is-author-date option seems to leave the author timestamp, and set the committer timestamp to be the same as the original … Read more
datetime.datetime.fromtimestamp() is correct, except you are probably having timestamp in miliseconds (like in JavaScript), but fromtimestamp() expects Unix timestamp, in seconds. Do it like that: >>> import datetime >>> your_timestamp = 1331856000000 >>> date = datetime.datetime.fromtimestamp(your_timestamp / 1e3) and the result is: >>> date datetime.datetime(2012, 3, 16, 1, 0) Does it answer your question? EDIT: … Read more