sql date format order dd-mm-yyyy - sql

Question

I am using this code to format date with SQL:
convert(VARCHAR, DT, 20)
The output is:
2016-01-01 20:40:12
But how do I get the day first.
I would like to get: dd-mm-yyyy:
01-01-2016 20:40:12
When I look in this table:
http://www.blackwasp.co.uk/sqldatetimeformats.aspx
I do not see this formatting.
Thx!

Answer 1

For dd/mm/yyyy hh:mi:ss
(CONVERT(varchar, DT, 103) + ' ' + CONVERT(varchar, DT, 108)) AS MyEuropeanDate
For dd-mm-yyyy hh:mi:ss
(CONVERT(varchar, DT, 105) + ' ' + CONVERT(varchar, DT, 108)) AS MyEuropeanDate
And to understand it right for the next time, look at the official doc that Gordon wisely linked in the comments :
https://msdn.microsoft.com/en-us/library/ms187928.aspx

Answer 2

Could not find a native tSQL function but you can do something like this:
DECLARE #Date DATETIME = '20160101';
SELECT CASE
WHEN LEN(CAST(DAY(#Date) AS VARCHAR)) = 1 THEN '0'+CAST(DAY(#Date) AS VARCHAR)
ELSE CAST(DAY(#Date) AS VARCHAR)
END
+'-'+
CASE
WHEN LEN(CAST(MONTH(#Date) AS VARCHAR)) = 1 THEN '0'+CAST(MONTH(#Date) AS VARCHAR)
ELSE CAST(MONTH(#Date) AS VARCHAR)
END
+'-'+
CAST(YEAR(#Date) AS VARCHAR)
+' '+
SUBSTRING(CONVERT(VARCHAR, #Date, 114), 1, 8);
OR you can do this:
DECLARE #Date DATETIME = '20160101';
SELECT (REPLACE(CONVERT(varchar, #Date, 103),'/','-') + ' ' + CONVERT(varchar, #Date, 108))
RESULT: