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 aorder 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 20104002.1.5 20105002.1.12 20112002.2.1 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 toowith 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：
cthe_value2.1.4 2142.1.5 2152.2.1 2212.1.12 222
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
