Question

I have a series of dates associated with a unique identifier in a table. For example:

1 | 1999-04-01 | 0000-00-00 | 0000-00-00 | 0000-00-00   | 2008-12-01 | 
2 | 1999-04-06 | 2000-04-01 | 0000-00-00 | 0000-00-00   | 2010-04-03 | 
3 | 1999-01-09 | 0000-00-00 | 0000-00-00 | 0000-00-00   | 2007-09-03 | 
4 | 1999-01-01 | 0000-00-00 | 1997-01-01 | 0000-00-00   | 2002-01-04 | 

Is there a way, to select the earliest date from the predefined list of DATE fields using a straightforward SQL command?

So the expected output would be:

1 | 1999-04-01
2 | 1999-04-06
3 | 1998-01-09
4 | 1997-01-01

I am guessing this is not possible but I wanted to ask and make sure. My current solution in mind involves putting all the dates in a temporary table and then using that to get the MIN()

thanks

Edit: The problem with using LEAST() as stated is that the new behaviour is to return NULL if any of the columns in NULL. In a series of dates like the dataset in question, any date might be NULL. I would like to obtain the earliest actual date from the set of dates.

SOLUTION: Used a combination of LEAST() and IF() in order to filter out NULL dates.

 SELECT LEAST( IF(date1=0,NOW(),date1), IF(date2=0,NOW(),date2), [...] );

Lessons learnt a) COALESCE does not treat ‘0000-00-00’ as a NULL date, b) LEAST will return ‘0000-00-00’ as the smallest value – I would guess this is due to internal integer comparison(?)

Answer 1

select id, least(date_col_a, date_col_b, date_col_c) from table

upd

select id, least (
  case when date_col_a = '0000-00-00' then  now() + interval 100 year else date_col_a end,
  case when date_col_b = '0000-00-00' then  now() + interval 100 year else date_col_b end) from table