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How to convert Big Int into Datetime style 109


How to convert Big Int into Datetime style 109

By : Marin Gömöri
Date : October 17 2020, 03:08 PM
I hope this helps you . How to convert the Big Int into datetime with style 109 that starts with mon dd yyyy hh:mi:ss:mmmAM , Example
code :
select ViaConvert = Convert(varchar(30),Convert(Datetime,((632979854880200000 - 599266080000000000) / 864000000000.0)),109)
      ,ViaFormat  = Format(Convert(Datetime,((632979854880200000 - 599266080000000000) / 864000000000.0)),'MMM dd yyyy hh:mm:ss tt')
ViaConvert                    ViaFormat
Nov  1 2006  1:44:48:017PM    Nov 01 2006 01:44:48 PM


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How to represent t-sql convert(datetime,data,style) in L2E query?

How to represent t-sql convert(datetime,data,style) in L2E query?


By : user3191914
Date : March 29 2020, 07:55 AM
I hope this helps you . I've found this: Convert String to Int in EF 4.0 It looks, like this trick might help, if we create a function to cast string into datetime.
UPDATE This problem left unsolved for me. As I needed a quick solution I converted the 'data' column to be of datetime type. This is not generic for future extensions, but it works for now. One of the solutions that are not really a solution.
Convert Datetime format to varchar format style workaround

Convert Datetime format to varchar format style workaround


By : anne
Date : March 29 2020, 07:55 AM
help you fix your problem You can use the style that most closely resembles what you want to compare on and use REPLACE to replace the - with \.
code :
SELECT REPLACE(CONVERT(<yourstyle>),'-','/');
SELECT REPLACE(CONVERT(varchar(23), yourdate, 121),'-','/');
Why does SQL Server convert VARCHAR to DATETIME using an invalid style?

Why does SQL Server convert VARCHAR to DATETIME using an invalid style?


By : user3556407
Date : March 29 2020, 07:55 AM
will be helpful for those in need Because the date is in a canonical format ie(20150101). The database engine falls over it implicitly. This is a compatibility feature.
If you swapped these around to UK or US date formats, you would receive conversion errors, because they cannot be implicitly converted.
code :
select convert(datetime,'20150425',99999999)
select convert(datetime,'20150425',100)
select convert(datetime,'20150425',113)
select convert(datetime,'20150425',010)
select convert(datetime,'20150425',8008135)
select convert(datetime,'20150425',000)
select convert(datetime2,'20150425',99999999)
datetime.datetime.strptime to convert a string to datetime it occurs the mistakes unconverted date remands: 2

datetime.datetime.strptime to convert a string to datetime it occurs the mistakes unconverted date remands: 2


By : swati
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You are overthinking this with Pandas. If your columns are appropriately named, you can feed directly to pd.to_datetime. In addition, avoid using Python's datetime module with Pandas:
code :
df = pd.DataFrame([[2015, 12, 20, 15, 10, 3.1234],
                   [2018, 5, 15, 10, 12, 65.432]],
                  columns=['year','month','day','hours','minutes','seconds'])

df['datetime'] = pd.to_datetime(df)

print(df)

   year  month  day  hours  minutes  seconds                   datetime
0  2015     12   20     15       10   3.1234 2015-12-20 15:10:03.123400
1  2018      5   15     10       12  65.4320 2018-05-15 10:13:05.432000
TypeError: an integer is required (got type datetime.datetime) when trying to convert a datetime object within a diction

TypeError: an integer is required (got type datetime.datetime) when trying to convert a datetime object within a diction


By : user3437040
Date : March 29 2020, 07:55 AM
this one helps. I see now what is happening. The line d = datetime.datetime(value) is the issue, you are passing a datetime object to the datetime.datetime() method which is why you get the valueError. This is unnecessary because you are already getting a datetime object out of your dictionary, so there is no reason to construct a new one. Here's the fixed code:
code :
import datetime

dic = {'Select start date': datetime.datetime(2019, 11, 7, 0, 0)}

for key, datetime_obj in dic.items():
     datetime_str = datetime_obj.strftime("%d %b %Y")
     print(datetime_str)
for key, value in dic.items():
    print(type(value))
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