unable to parse String date "2019-02-23T13:59:13.312401" - parsing

I am trying for many hours to parse this date String which is returned by Shodan. "2019-02-23T13:59:13.312401" ISO format is throwing exception.
java.lang.IllegalArgumentException: Invalid format: "2019-02-14T10:16:35.313860" is too short
at org.joda.time.format.DateTimeFormatter.parseDateTime(DateTimeFormatter.java:945)
I am using Joda DateTime library with Google Gson. Anyone knows which format is this and how can I convert it to DateTime object from String.?

Shodan stores the timestamps in ISO 8601 format which you can read more about here:
https://docs.python.org/2/library/datetime.html#datetime.datetime.isoformat
And see here for a previous answer on parsing the format using Joda time:
Converting ISO 8601-compliant String to java.util.Date

Related

how to convert the date format from dd-mmm-yyyy to yyyy-mm-dd in ruby on rails

I am uploading the csv file. The csv file contains the date format in dd-mm-yyyy format. How can I change it to yyyy-mm-dd format. I also tried to_date function but it gives the error as invalid date.
You can do this by below code
require 'date'
date = DateTime.parse("22-02-2019")
formatted_date = date.strftime('%Y-%m-%d')
output
"2019-02-22"
please go through with this document https://www.rubydoc.info/stdlib/core/Time:strftime you will find out the different formatting style

Change datetime format from one to another programmatically

In my program, I need to change the input datetime information from Turkish (d.M.yyyy) language to English (M/dd/yyyy).
When client edits datetime information in Turkish datetime format(For Example: 24.9.2015), i need to convert this information in English datetime format(Like; 9/24/2015).
I'm trying to parse the input, but not convert the correct format that i need, code block as in the following.
DateTime.TryParseExact(InputDate,"d.M.yyyy", CultureInfo.InvariantCulture, DateTimeStyles.None, out result))
I need the output named "result" in English U.S datetime format. I also tried to change the code as CultureInfo(en-US) but it doesn't work.
Any ideas?
In what language are you trying to achieve this?
Assuming it is c#, DateTime does not have a specific format until you try to convert it into a string and you can convert it to US version by specifying the correct format when calling ToString()
string americanFormat = myDate.ToString("MM/dd/yyyy");

Date format in Talend

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 : 24/12/2013 16:25:47.328000000
I have tried:
TalendDate.parseDate("yyyy-MM-dd'T'HH:mm:ss.SSSXXX",row1.REGISTERED_ON,true),
TalendDate.parseDate("yyyy-MM-dd'T'HH:mm:ss",row1.REGISTERED_ON,true)
but still I get that same error.
Can anyone help shed some light please?
I'm not familiar with Talend, but you say
My date format is likewise : 24/12/2013 16:25:47.328000000
and
I have tried
TalendDate.parseDate("yyyy-MM-dd'T'HH:mm:ss.SSSXXX",row1.REGISTERED_ON,true),
TalendDate.parseDate("yyyy-MM-dd'T'HH:mm:ss",row1.REGISTERED_ON,true)
but the date you specify is in dd/MM/yyyy format. I imagine that's the cause of your problems.
I am Able to parse given string date using below format, below statement should work for you,
Input String "24/12/2013 16:25:47.328000000"
Format: "dd/MM/yyyy HH:mm:ss.SSSS"
System.out.println(TalendDate.parseDate("dd/MM/yyyy HH:mm:ss.SSSS", "24/12/2013 16:25:47.328000000"));
See more formats on "How to parse String to Date"
to manage date in talend what i prefer to do is
1) set data type of your_date_column to string in your input file
2) parse your_date_column at the time of mapping using
new SimpleDateFormat("yyyy-MM-dd HH:mm:ss").parse(your_date_column)
3) set data type of our output_date_column to Date with date pattern :- "yyyy-MM-dd HH:mm:ss"
Hope it will Work :)

Convert SPSS string variable to numeric and date

I have the date variable which is in string format. I need to convert this to numeric or date format. How can I do this?
Example:
var1
2010/07/09
2010/07/16
Thanks,
Tanuvi
If you read in the date value as a string, you can use Transform > Date and Time Wizard to convert it to an SPSS date variable. Although date variables look like dates in various formats, they are actually represented as the number of seconds since 1582 and can consist of calendar and time information.

Need conversion technique from Times ten dateTime to Saxon Timestamp

Trying to retrieve a dateTime Information from Timesten Database and use it in Saxon Xquery .below is the example for that and getting the below error . Do we need to convert timesten dateTime to saxom dataTime if yes how to do that ? pls help me if have idea.
let $DateVar:=fn:data($PERSON/BIRTHDAY)
where as $PERSON/IN_BIRTHDAY is 2010-04-04 03:16:04.000000
if I am trying
let $day-b-DT :=day-from-dateTime($DateVar)
I am getting
Validation error
FORG0001: Invalid dateTime value "2010-04-04 03:16:04.000000" (Day must be two digits)
net.sf.saxon.s9api.SaxonApiUncheckedException: Invalid dateTime value "2010-04-04 03:16:04.000000" (Day must be two digits)
I believe the problem is just your string format, which should be
"2010-04-04T03:16:04.000000". See the documentation for dateTime for more information.
I don't know anything about the Times ten database, or whether you retrieve values in a "rich" format which you happen to be formatting to a string (in which case you should be able to specify a different format) but I believe that's what's wrong.

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