Convert an Int to a Date Field

Convert INT to DATETIME (SQL)

you need to convert to char first because converting to int adds those days to 1900-01-01

select CONVERT (datetime,convert(char(8),rnwl_efctv_dt ))

here are some examples

select CONVERT (datetime,5)

1900-01-06 00:00:00.000

select CONVERT (datetime,20100101)

blows up, because you can't add 20100101 days to 1900-01-01..you go above the limit

convert to char first

declare @i int
select @i = 20100101
select CONVERT (datetime,convert(char(8),@i))

Convert an Int to a date field

Simple cast as date could work

Select cast(cast(20161011 as varchar(8)) as date)

Returns

2016-10-11

If your data is suspect, you could also use Try_Convert()

Select Try_Convert(date,cast(2610 as varchar(8)))

Returns

NULL

CONVERT Date INT to DATE in SQL

(Read below for my answer for if it's an int column)

Assuming it's a textual string:

Assuming that datadate is a string (character, text, etc) column and not a date/datetime/datetime2/datetimeoffset column, then use the CONVERT function with style: 23. The 23 value corresponds to ISO 8601 because the values are in yyyy-MM-dd-order, even though they're missing dashes.

This page has a reference of style numbers: https://learn.microsoft.com/en-us/sql/t-sql/functions/cast-and-convert-transact-sql?view=sql-server-ver15

SELECT
    *
FROM
    (
        SELECT
            myTable.*
            CONVERT( date, datadate, 23 ) AS valueAsDate
        FROM
            myTable
    ) AS q
WHERE
    q.valueAsDate = DATEADD( dd, -1, GETDATE() )

Assuming it's an actual int column:

The quick-and-dirty way is to convert the int to varchar and then use the same code as above as if it were a textual field - but don't do this because it's slow:

SELECT
    *
FROM
    (
        SELECT
            myTable.*,
            CONVERT( char(8), datadate ) AS valueAsChar,
            CONVERT( date, CONVERT( char(8), datadate ), 23 ) AS valueAsDate
        FROM
            myTable
    ) AS q
WHERE
    q.valueAsDate = DATEADD( dd, -1, GETDATE() )

Assuming it's an actual int column (better answer):

We'll need to use DATEFROMPARTS and extract each component using Base-10 arithmetic (fun)!

If we have an integer representing a formatted date _{(the horror)} such as 20200531 then:

• We can get the day by performing MOD 31 (e.g. 19950707 MOD 31 == 7)
• We can get the month by first dividing by 100 to remove the day part, and then MOD 12: (e.g. 20200531 / 100 == 202005, 202005 MOD 12 == 5)
• We can get the year by dividing by 10,000, (e.g. 20200531 / 10000 == 2020).

Btw:

• SQL Server uses % for the Modulo operator instead of MOD.
• Integer division causes truncation rather than producing decimal or floating-point values (e.g. 5 / 2 == 2 and not 2.5).

Like so:

SELECT
    q2.* 
FROM
    (
        SELECT
            q.*,
            DATEFROMPARTS( q.[Year], q.MonthOfYear, q.DayOfMonth ) AS valueAsDate
        FROM
            (
                SELECT
                    myTable.*,

                    ( datadate % 31 ) AS DayOfMonth,
                    ( ( datadate / 100 ) % 12 ) AS MonthOfYear,
                   ( datadate / 10000 ) AS [Year]
                FROM
                    myTable
            ) AS q
    ) AS q2
WHERE
    q2.valueAsDate = DATEADD( dd, -1, GETDATE() )

Obviously, having two nested subqueries is a pain to work with _{(SQL has terrible ergonomics, I don't understand how or why SQL doesn't allow expressions in a SELECT clause to be used by other expressions in the same query - it's really bad ergonomics...)} - but we can convert this to a scalar UDF (and SQL Server will inline scalar UDFs so there's no performance impact).

This function has a TRY/CATCH block in it because of the possibility that you process an invalid value like 20209900 (which isn't a real date as there isn't a 99th month with a 0th day in 2020). In this event the function returns NULL.

CREATE FUNCTION dbo.convertHorribleIntegerDate( @value int ) RETURNS date AS
BEGIN

    DECLARE @dayOfMonth  int = @value % 31;
    DECLARE @monthOfYear int = ( @value / 100 ) % 100;
    DECLARE @year        int = @value / 10000;

    BEGIN TRY
        RETURN DATEFROMPARTS( @dayOfMonth, @monthOfYear, @year );    
    END TRY;
    BEGIN CATCH
        RETURN NULL;
    END CATCH;

END

Which we can use in a query like so:

SELECT
    myTable.*,
    dbo.convertHorribleIntegerDate( datadate ) AS valueAsDate
FROM
    myTable

As SELECT cannot share expression results with other expressions in the same query, you'll still need to use an outer query to work with valueAsDate (or repeat the dbo.convertHorribleIntegerDate function call):

SELECT
    *
FROM
    (
        SELECT
            myTable.*,
            dbo.convertHorribleIntegerDate( datadate ) AS valueAsDate
        FROM
            myTable
    ) AS q
WHERE
    q.valueAsDate = DATEADD( dd, -1, GETDATE() )

How to convert a column of type integer to type datetime in python?

Just apply pd.to_datetime directly to (the string conversion of) the column, no need to use string slicing here:

Freshman['ADMIT_DATE'] = pd.to_datetime(Freshman['ADMIT_DATE'].astype(str), format='%Y%m')

There is no requirement for there to be a delimiter between the digits:

>>> import pandas as pd
>>> df = pd.DataFrame({'ADMIT_DATE': [200110, 201604]})
>>> df['ADMIT_DATE'] = pd.to_datetime(df['ADMIT_DATE'].astype(str), format='%Y%m')
>>> df
  ADMIT_DATE
0 2001-10-01
1 2016-04-01
>>> df.dtypes
ADMIT_DATE    datetime64[ns]
dtype: object

Int to date in SQL

This works for me in SQL Server 2017:

select cast(cast(20190430 as nvarchar) as DATE) AS DateResult

Output:

DateResult
2019-04-30

Convert Integer field Date time to Date Time

You can use DATEADD to convert UNIX timestamp to DATETIME and FORMAT function to format it:

SELECT FORMAT(DATEADD(SECOND, 1512543210, '19700101'), 'dd-MM-yyyy hh:mm:ss')
-- 06-12-2017 06:53:30

Having said that, Java has DateTimeFormatter class for formatting dates. And timestamps could be used to construct date objects directly.

How to convert integer into date object python?

I would suggest the following simple approach for conversion:

from datetime import datetime, timedelta
s = "20120213"
# you could also import date instead of datetime and use that.
date = datetime(year=int(s[0:4]), month=int(s[4:6]), day=int(s[6:8]))

For adding/subtracting an arbitary amount of days (seconds work too btw.), you could do the following:

date += timedelta(days=10)
date -= timedelta(days=5)

And convert back using:

s = date.strftime("%Y%m%d")

To convert the integer to a string safely, use:

s = "{0:-08d}".format(i)

This ensures that your string is eight charecters long and left-padded with zeroes, even if the year is smaller than 1000 (negative years could become funny though).

Further reference: datetime objects, timedelta objects

convert integer to date in python

You can represent dates with the .dt method of the Series

Adapted from my similar answer https://stackoverflow.com/a/66038817/4541045

import pandas as pd
df = pd.DataFrame({"dates": [200101, 200202, 202010]})
#     dates
# 0  200101
# 1  200202
# 2  202010
df.dates = pd.to_datetime(df.dates, format="%Y%m")  # make a datetime
#        dates
# 0 2001-01-01
# 1 2002-02-01
# 2 2020-10-01
df.dates.dt.strftime("%Y%m")  # represent datetime Series
# 0    200101
# 1    200202
# 2    202010
# Name: dates, dtype: object

NOTE df.dates is just referring to the column named dates in this case

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