Question

I need to find a way to get the data with the highest versionNumber.

Here is my database design:

VERSIONNUMBER - varchar(15)
DOWNLOADPATH - varchar(100)

Lets say I have records like:

VERSIONNUMBER -------- DOWNLOADPATH
1.1.2                  a.com
1.1.3                  b.com
2.1.4                  c.com
2.1.5                  d.com
2.2.1                  e.com

I need to get the record with the versionnumber 2.2.1. Need some help with the sql though 🙂

Thank you for any help

Answer 1

Try this:

with a as
(
    select * from (values
    ('1.1.2'),('1.1.3'),('2.1.4 '), ('2.1.5'), ('2.2.1') ) as b(c)
)
select c, PARSENAME(c,1),PARSENAME(c,2), PARSENAME(c,3)
from a
order by 
convert(int,PARSENAME(c,3)),
convert(int,PARSENAME(c,2)),
convert(int,PARSENAME(c,1))

Inspired from: http://www.sql-server-helper.com/tips/sort-ip-address.aspx

with a as
(
    select * from (values
    ('1.1.2'),('1.1.3'),('2.1.4 '), ('2.1.5'), ('2.2.1') ) as b(c)
),
x as 
(
    select c, 
       convert(int,PARSENAME(c,3)) * 100 
       + convert(int,PARSENAME(c,2)) * 10 
       + convert(int,PARSENAME(c,1)) * 1 as the_value
    from a
)
select c from x where the_value = (select MAX(the_value) from x)

In software development, it is typical to find a minor version number that has two digits in it, the version’s number don’t have any bearing with number’s value, thus version 1.12 is greater than 1.5; to compensate for that, you must pad the digits adequately:

    -- Use this, the query above is not future-proof :-)
with a as
(
    select * from (values
    ('2.1.4 '), ('2.1.12'), ('2.1.5'), ('2.2.1') ) as b(c)
),
x as 
(
    select c, 
       convert(int,PARSENAME(c,3)) * 100*100*100 
       + convert(int,PARSENAME(c,2)) * 100*100 
       + convert(int,PARSENAME(c,1)) * 100 as the_value
    from a
)
select c, the_value from x   
order by the_value

Output:

2.1.4   2010400
2.1.5   2010500
2.1.12  2011200
2.2.1   2020100

If you don’t take that into consideration(as with the following query):

with a as
(
    select * from (values
    ('2.1.4 '), ('2.1.12'), ('2.1.5'), ('2.2.1') ) as b(c)
),
x as 
(
    select c, 
       convert(int,PARSENAME(c,3)) * 100
       + convert(int,PARSENAME(c,2)) * 10
       + convert(int,PARSENAME(c,1)) * 1 as the_value
    from a
)
select c, the_value from x   
order by the_value;


    -- KorsG's answer has a bug too
with a as
(
    select * from (values
    ('2.1.4 '), ('2.1.12'), ('2.1.5'), ('2.2.1') ) as b(c)
),
x as 
(
    select c, 
       CAST(REPLACE(c, '.', '') AS int) as the_value
    from a
)
select c, the_value from x   
order by the_value      

Those two queries will yield the same (incorrect) output:

c           the_value
2.1.4   214
2.1.5   215
2.2.1   221
2.1.12  222

The 2.2.1 and 2.1.12’s value overlapped. That also happens when you merely remove the dots and directly convert the resulting string to int. 2.1.12 become two thousand one hundred twelve, 2.2.1 become two hundred twenty one. 2.2.1 is greater than 2.1.12, not less than