I am inserting dates that look like:
'19APR2014:08:42:32.123456'
I am interpreting their format as
'DDMONYYYY:HH24:MI:SS.FFFFFF'
Though I have not seen any times after 12:59:59 I am assuming a 24-hour clock. Hive does not seem to understand what I want to do:
HiveException: Error evaluating unix_timestamp(date_string,'DDMONYYYY:HH24:MI:SS.FFFFFF')
Any ideas what I am doing wrong or what might be wrong with my format string?
Have you tried ddMMMyyyy:HH:mm:ss.SSS? According to Hive manual a pattern string in function unix_timestamp(string date, string pattern) should comply to Java's SimpleDateFormat(see manual and javadocs).
Related
How can I represent format of 2018-01-01T17:11:11.111+06:00 in string?
I've tried yyyy-MM-dd'T'HH:mm:ss.SSS+HH:mm, but it didn't work.
It is needed for range query in elasticsearch using elastic4s.
Let' try:
YYYY-MM-dd'T'HH:mm:ss.SSSz
I suggest to you this app for check the format:
https://esddd.herokuapp.com/
I have been looking at the above question and have most of it correct.
I am going to get a datetime in Zulu, and then will want to output that format.
My first go is just as simple as:
DateFormat format = new DateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'");
My issue I am having is the T and Z. Obviously T is used to separate the date from the time and the Z is representative of Zulu time.
That being said the users will be entering a datetime in Zulu, so it wont need to be converted from Local to Zulu, so i was not sure if 'Z' is an acceptable result. I was not sure if there is a different want to handle this, or if my result was the best answer.
Try this package, Jiffy.
String isoFomart = Jiffy().format(); // This will return ISO format from now
You can also add your DateTime object
String isoFomart = Jiffy(DateTime.now()).format(); // This will also return ISO format from now
Hope this helped
The DateTime object has a method called: toIso8601String which is used to return an ISO formatted string. The 'Z' will be added if isUTC is true, otherwise the result will not have the Z in it.
Make sure that the DateTime object itself is correctly set to UTC as if you look in the constructor for the class will tell you a lot of the defaults are local with the exception of the DateTime.utc() static function.
In that concept, you dont really need a DateFormat use to define an iso string.
I am facing an issue with Talend dates. I have tried several solutions but still an "unparseable date" error persists.
My date format is of the form : "2006-05-27 17:00:00.000"
Can you help me ?
you can use below talendDate function to parse your string into date..
TalendDate.parseDate("yyyy-MM-dd HH:mm:ss.sss","2006-05-27 17:00:00.000")
this would take input as string and return you date.
If you don't handle the conversion yourself in a tMap but just want to use a schema, then: In your mapping configuration in the date field, you can add the following string:
yyyy-MM-dd HH:mm:ss.SSS
to set the correct format mapping for the date string. Otherwise the answer of garpitmzn is the way to go.
you should use this function in talend to fetch date:
TalendDate.parseDate("yyyy-MM-dd HH:mm:ss.SSS","2016-01-12 12:45:00.000")
I am trying to get this pattern 'dd-MM-yyyy' with a variable of type DateTime
#{DateTimeFormat.forPattern("dd-MM-YYYY").parseDateTime(user.birthday.toString)}
But I am getting this error
java.lang.IllegalArgumentException: Invalid format: "2015-12-10T00:00:00.000Z" is malformed at "15-12-10T00:00:00.000Z"
Is there a way to do this with nscala-time?
makes a difference if I am using UTC?
UPDATE
For the moment I am casting to Date and doing this
#{Dates.format(user.birthday.toDate, "dd-MM-YYYY")}
But maybe is a better way without casting
thank you
So, if I understood your question correctly, you are trying to achieve the following:
Parse date from a string using a date format
Print/Display the date in another format.
Try the below:
#{Dates.format(DateTimeFormat.forPattern("yyyy-MM-dd'T'HH:mm:ss.SSSZ").parseDateTime(user.birthday.toString), "dd-MM-YYYY")}
I have a date string in the format "2013-01-31T10:10:05.000Z". I want to convert this string to a Date object in extjs.
I have tried to use Ext.Date.parse("2013-01-31T10:10:05.000Z","Y-m-dTH:i:s.uZ"). But it is returning undefined.
I also tried with new Date("2013-01-31T10:10:05.000Z"), but it is also returning undefined.
Note: I have tried in IE8 browser.
Could anyone please help me to convert the above date string to Date object?
Thanks a lot sra. Now I am getting the result as ...UTC+5:30... Is there any way to convert this in IST format?
Try Ext.Date.parse("2013-01-31T10:10:05.000Z","c");
The c is the format type for ISO 8601 formatted dates
See the Ext.Date API for more details on this or other available formats
That's because 'T' and 'Z' are special characters in the Date format: http://docs.sencha.com/extjs/4.2.1/#!/api/Ext.Date
You have to escape them like this: Ext.Date.parse("2013-01-31T10:10:05.000Z","Y-m-d\\TH:i:s.u\\Z")