求教一个oracle的分组求和问题

1、最简单的方法，使用 UNION SELECT '苹果' AS NAME, SUM(WEIGHT) AS WEIGHT FROM ZZZZ_TTTT WHERE NAME LIKE '%苹果'UNION ALL SELECT '梨' AS NAME, SUM(WEIGHT) AS WEIGHT FROM ZZZZ_TTTT WHERE NAME LIKE '%梨'2、使用 CASE WHEN SELECT NAME, SUM(WEIGHT)FROM (SELECT CASE WHEN IN...

一般的分组求和：select t.column_name,count(*) from table_name t group by t.column_name

group by分组+sum求和就行啊，根据你的表格可能需要用到substr截取。另外，建议使用时长不要带单位，单位可以另存或者标注。具体的是 select 程序名称,sum(使用时长) from table group by 程序名称 如果不行，那么就把sum(使用时长)替换成sum(substr(使用时长,1,length(使用时长)-1))||'s'...

sum(统计字段名) over(partition by 分组字段1),sum(统计字段名) over(partition by 分组字段2),sum(统计字段名) over(partition by 分组字段3),...sum(统计字段名) over(partition by 分组字段n)from t ;

select nvl(job, '总计：'), count(1) 员工数量, count(distinct deptno) 部门数量 from (select deptno, nvl(job, ' ') job from emp) a group by rollup(job) order by job以上是oracle语法，你试下。

select decode(t.f_disqua_level,'1','A类','2','B类','3','C类','4','D类','') as 等级, t.f_process as 发现地点, t.f_discovery_time as 发现时间, t.f_description as 不合格描述, b.f_dutydept as 责任部门, b.f_assess_multiple as 考核倍数, b.f_assess_...

SELECT b.job, SUM(A.SAL)FROM staff a, staff b WHERE a.job <= b.job GROUP BY b.job ORDER BY b.job

WITH t1 AS(SELECT 'a' c1, 2 c2 FROM dual UNION ALL SELECT 'a' c1, 4 c2 FROM dual UNION ALL SELECT 'b' c1, 1 c2 FROM dual UNION ALL SELECT 'b' c1, 8 c2 FROM dual )SELECT c1, SUM(c2), SUM(c2)/(SELECT SUM(c2) FROM t1) FROM t1 GROUP BY c1 ...

select a1,count(*) from (select a1 from x union all select a2 from x ) group by a1 order by a1;

将相同字段的值加在一起需要用sum函数。如emp表中有如下数据：现要按deptno为分组，将sal的值加在一起，可用如下语句：select deptno,sum(sal) sal from emp group by deptno;查询结果：