1. select sum(a) as c,sum(b) as d,(isnull(sum(a),0)+isnull(sum(b),0)) as e from table2.select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(b,0)) as e from table
--要考虑NULL值对结果的影响 select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(d,0)) as e from table
--或者这样写:select c,d,isnull(a,0)+isnull(d,0) as e from(select sum(a) as c,sum(b) as d from table)a
select c,d,(c+d) as e from (select sum(a) as c,sum(b) as d from table) A
select sum(a) as c,sum(b) as d,(sum(a)+sum(b)) as e from table
select语句查询的只能是表中已有的字段 所以应该是 select sum(a) as c,sum(b) as d,(sum(a)+sum(b)) as e from table;
试试这样;select c,d,e as 'c+d' from (select sum(a) as c,sum(b) as d,(isnull(sum(a),0)+isnull(sum(b),0)) as e from tableName)as t1
要考虑NULL值对结果的影响 1. select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(d,0)) as e from table 2. select c,d,isnull(c,0)+isnull(d,0) as e from(select sum(a) as c,sum(b) as d from table)a 1,2我觉得执行结果肯定是一样的,但是,是否1比2会更加有效率呢
1. select sum(a) as c,sum(b) as d,(isnull(sum(a),0)+isnull(sum(b),0)) as e from table2.select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(b,0)) as e from table
select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(d,0)) as e from table
from(select sum(a) as c,sum(b) as d from table)a
(select sum(a) as c,sum(b) as d from table) A
所以应该是
select sum(a) as c,sum(b) as d,(sum(a)+sum(b)) as e from table;
1.
select sum(a) as c,sum(b) as d,sum(isnull(a,0)+isnull(d,0)) as e from table
2.
select c,d,isnull(c,0)+isnull(d,0) as e
from(select sum(a) as c,sum(b) as d from table)a
1,2我觉得执行结果肯定是一样的,但是,是否1比2会更加有效率呢