本文参考网址:http://www.itpub.net/thread-1020772-1-1.html 我们在日常的工作中,经常遇到了针对某一列的值,进行求和,求平均值,在一些特殊的业务场景下,我们需要对某一列进行求积操作,那我们该如何实现呢,下面先介绍,我 们对字符串的求和操作. 针对字符串的求和操作: 1.有分隔符的字符串: SELECT STR, SUM(STR) OVER(ORDER BY LV ASC) AS RS , LEVEL) AS STR, LEVEL AS LV FROM D
oracle sum()over函数的使用 over不能单独使用,要和分析函数:rank(),dense_rank(),row_number()等一起使用. over函数的参数:over(partition by columnname1 order by columnname2) 含义,按columname1指定的字段进行分组排序,或者说按字段columnname1的值进行分组排序. 例如:employees表中,有两个部门的记录:department_id ＝10和20 select depa
//将当前行某列的值与前面所有行的此列值相加,即累计求和: //方法一: with t as( select 1 val from dual union all select 3 from dual union all select 5 from dual union all select 7 from dual union all select 9 from dual) select val, sum(val) ov
import java.util.Stack; /** * * Source : https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/ * * * Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. * An example is the root-to-leaf path 1->2
select s.slice_date_to ,s.made_id ,sum(s.steup_count)over(partition by s.made_id order by s.slice_date_to) from rpt_sliceresult_customer_setup s where s.group_type='SECOND_PRODUCE_DATE' and s.evaluate_id='CQM20150818112644734'; select s.slice_dat
select prod.amount,prod.plansum,(prod.plansum-prod.amount) as borrow,d.enum_value from ----结果集相减(select t.Quo_Prod_List_Price * t.QUO_PROD_VOLUME as amount ,----列值相乘 t.quo_prod_plan_cost * t.quo_prod_volume as plansum , t.broad_class as broad_c
第一种: select sum(decode(count1,null,0,count1) +decode(count2,null,0,count2) +decode(count3,null,0,count3)) a from test_table 第二种: select sum (case when count1 is not null then count1 when count2 is not null then count2 when count3 is not null then cou
1.--dense_rank()分析函数(查找每个部门工资最高前三名员工信息) select * from (select deptno,ename,sal,dense_rank() over(partition by deptno order by sal desc) a from scott.emp) where a<=3 order by deptno asc,sal desc ; 结果: --rank()分析函数(运行结果与上语句相同) select * from (select dep
