SQL将日期dmmyyyy和ddmmyy转换为yyyy-mm-dd

Question

I have the following problem. I have some dates with the following format '15122019' and I need it in this format 2019-12-15, which I already solved it in the following way.

select convert (date, Stuff(Stuff('15122018',5,0,'.'),3,0,'.'),104)

The real problem is when the dates come like this '3122019' the conversion can not be done because the length is shorter. Is ther e another way to do it? I've been trying to solve it for several hours. And another question, can this query be parameterized?

Answer 1

Try this:

 DECLARE @date VARCHAR(20)
 SET @date ='3122019'

 IF(LEN(@date) = 8)
 BEGIN  
    SET @date = Stuff(Stuff(@date,5,0,'.'),3,0,'.');
    SELECT CONVERT(DATE, @date , 103);
 END
 ELSE IF(LEN(@date) = 7) 
 BEGIN

    SET @date = Stuff(Stuff(@date,4,0,'.'),2,0,'.');
    IF(ISDATE(@date)=1)
    BEGIN
        SELECT CONVERT(DATE, @date , 103);          
    END    
    ELSE
    BEGIN
        SET @date = Stuff(Stuff(@date,4,0,'.'),3,0,'.');    
            SELECT CONVERT(DATE, @date , 103);
    END
 END
 ELSE IF(LEN(@date) = 6)
 BEGIN
     SET @date = Stuff(Stuff(@date,3,0,'.'),2,0,'.');
     SELECT CONVERT(DATE, @date , 103);
 END

Answer 2

Such conversation can be achieved by:

• Casting integer value to DATE using intermediate FORMAT transformation to a recognizable for conversation string pattern.
  • style 105 applied to match the input as dd-mm-yyyy
  • style 05 to match dd-mm-yy

SQL:

-- input format: dmmyyyy
SELECT CONVERT(DATE, FORMAT(3012019, '##-##-####'), 105)
-- result: 2019-01-03

-- input format: dmmyy
SELECT CONVERT(DATE, FORMAT(30119, '##-##-##'), 05)
-- result: 2019-01-03

This will work fine with a single (and double) digit day number, however, it indeed requires a double-digit month

Answer 3

You can add 0 to the left and take 8 chars with right . like RIGHT('0'+'15122018',8) . it work with 15122018 and 3122018

select convert (date, Stuff(Stuff( RIGHT('0'+'15122018',8) ,5,0,'.'),3,0,'.'),104)