我能否仅使用一次表扫描就获得至少2列大于给定值的列

Question

This is my example data (there are no indexes and I do not want to create any):

CREATE TABLE tblTest ( a INT , b INT );
INSERT INTO tblTest ( a, b ) VALUES
( 1 , 2 ),
( 5 , 1 ),
( 1 , 4 ),
( 3 , 2 )

I want the minimum value in of both column a and column b which is greater then a given value. Eg if the given value is 3 then I want 4 to be returned.

This is my current solution:

SELECT MIN (subMin) FROM 
(
 SELECT MIN (a) as subMin FROM tblTest
 WHERE a > 3 -- Returns 5
 UNION 
 SELECT MIN (b) as subMin FROM tblTest
 WHERE b > 3 -- Returns 4
) 

This searches the table twice - once to get min(a) once to get min(b) .

I believe it should be faster to do this with just one pass. Is this possible?

Answer 1

You want to use conditional aggregatino for this:

select min(case when a > 3 then a end) as minA,
       min(case when b > 3 then b end) as minB
from tblTest;

To get the minimum of both values, you can use a SQLite extension, which handles multiple values for min() :

select min(min(case when a > 3 then a end),
           min(case when b > 3 then b end)
          )
from tblTest

The only issue is that the min will return NULL if either argument is NULL. You can fix this by doing:

select coalesce(min(min(case when a > 3 then a end),
                    min(case when b > 3 then b end)
                   ),
                min(case when a > 3 then a end),
                min(case when b > 3 then b end)
               )
from tblTest

This version will return the minimum value, subject to your conditions. If one of the conditions has no rows, it will still return the minimum of the other value.

Answer 2

From the top of my head, you could modify the table and add a min value column to store the minimum value of the two columns. then query that column.

Or you can do this:

select min(val)
from
(
    select min(col1, col2) as val
    from table1
)
where 
val > 3

The outer SELECT, queries the memory, not the table itself.

Check SQL Fiddle