Pass mm/dd/yyyy to yyyy-mm-dd date format - sql

Question

I would like to know if there is a function or a simple way to pass the format date mm/dd/yyyy
For example: 11/01/2022 (November 1st 2022)
To this format yyyy-mm-dd
For example: 2022-11-01
Nowadays, I am slicing the date as a string, getting the day, month and year individually and put them together as I want, but I'd like to know if there is a better way to do it.

Answer 1

This can be done using PARSE which has the [USING culture] option:
PARSE ( string_value AS data_type [ USING culture ] )
In your example
SELECT PARSE('11/01/2022' AS datetime2 USING 'en-US');
Output:
2022-11-01
The above gives you date type which should be good for most tasks but if you need to get a string in a preferred format you can use FORMAT.
A. Culture specific output:
DECLARE #d DATE = PARSE('11/01/2022' AS datetime2 USING 'en-US');
SELECT FORMAT( #d, 'd', 'en-US' ) 'US English'
,FORMAT( #d, 'd', 'en-gb' ) 'British English'
,FORMAT( #d, 'd', 'en-au' ) 'Australian'
,FORMAT( #d, 'd', 'de-de' ) 'German'
,FORMAT( #d, 'd', 'zh-cn' ) 'Chinese Simplified (PRC)';
US English British English Australian German Chinese Simplified (PRC)
11/1/2022 01/11/2022 1/11/2022 01.11.2022 2022/11/1
(Please note some countries always use two digits for day and month and some don't)
B. Fixed format output:
DECLARE #d DATE = PARSE('11/01/2022' AS date USING 'en-US');
SELECT FORMAT( d, 'yyyy-MM-dd') 'yyyy-MM-dd';
2022-11-01

Answer 2

Assuming you're storing a date in a TEXT field? That is bad, but if that's indeed the case, the first step is to first parse it into a DATETIME object:
PARSE('11/01/2022' AS datetime USING 'en-US')
Now that it's in a DATETIME, use FORMAT() to format it to your desired format:
SELECT FORMAT( PARSE('11/01/2022' AS datetime USING 'en-US') , 'yyyy-MM-dd')
Returns:
2022-11-01