拆分字符串并在mssql中返回最大

Question

我需要找到一种方法来获取具有最高versionNumber的数据。

这是我的数据库设计：

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

可以说我有以下记录：

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

我需要获取版本号为2.2.1的记录。需要一些帮助的SQL :)

感谢您的任何帮助

Answer 1

试试这个：

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))

灵感来自：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)

在软件开发中，通常会找到一个包含两个数字的次要版本号，该版本号与该数字的值没有任何关系，因此版本1.12大于1.5；否则，版本号不大于1.5。为了弥补这一点，您必须适当
填充 数字：

    -- 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

输出：

2.1.4   2010400
2.1.5   2010500
2.1.12  2011200
2.2.1   2020100

如果您不考虑这一点（与以下查询一样）：

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

这两个查询将产生相同（不正确）的输出：

c           the_value
2.1.4   214
2.1.5   215
2.2.1   221
2.1.12  222

2.2.1和2.1.12的值重叠。当您只删除点并将结果字符串直接转换为int时，也会发生这种情况。2.1.12变成212，2.2.1变成212。2.2.1大于2.1.12，不小于