oracle 默认排序空值在后面，显示在前面用nvl(column,-1) 或nulls first 与之对应的是nulls last
取手机号码的后四位
substr(phoneNum,-4)
日期
trunc(date,'yy') 年初
trunc(date,'mm') 月初
with 用法
with类似创建一张临时表
with e as (select rownum sn,dept_code,job from v)
select dept_code from e;
正则表达式
regexp_replace
regexp_count
select regexp_replace(data,'[^0-9]','') dname from dept;
注意^的位置在方括号内，所有字符之前，是除[0123456789]的意思，如果不在方括号内，则表示字符串开始的位置'^hell'
select id,case when trx='PY' then '取款' else '存款' end 存取类型,amt 金额,
sum(case when trx='PY' then -amt else amt end) over(order by id)
from v order by id;
select deptno,empno,ename,sal,first_value(ename) over(partition by deptno order by empno) as 工资最高的人min,
first_value(ename) over(partition by deptno order by empno desc) as 工资最高的人max
from emp
where deptno=20
order by 1,4 desc;
with t1 as
(select grouping(t.b) as gp_b,
t.b as 品牌,
sum(case t.a when '门店1' then t.c end) as 销量_门店1,
sum(case t.a when '门店2' then t.c end) as 销量_门店2,
sum(case t.a when '门店3' then t.c end) as 销量_门店3,
sum(t.c) as 销量_合计,
sum(case t.a when '门店1' then t.d end) as 收入_门店1,
sum(case t.a when '门店2' then t.d end) as 收入_门店2,
sum(case t.a when '门店3' then t.d end) as 收入_门店3,
sum(t.d) as 收入_合计
from T t
group by rollup(t.b)
order by 1 desc ,2)
select case when gp_b=1 then '销量合计' else 品牌 end as 品牌,
销量_门店1 as 门店1,
销量_门店2 as 门店2,
销量_门店3 as 门店3,
销量_合计 as 合计
from t1
union all
select case when gp_b=1 then '收入合计' else 品牌 end as 品牌,
收入_门店1 as 门店1,
收入_门店2 as 门店2,
收入_门店3 as 门店3
收入_合计 as 合计
from t1
union all
select case when gb_b=1 then '收入合计' else 品牌 end as 品牌，
round(销量_门店1*收入_门店1/销量_合计,2) as 门店1,
round(销量_门店2*收入_门店2/销量_合计,2) as 门店2,
round(销量_门店3*收入_门店3/销量_合计,2) as 门店3,
收入_合计 as 合计
from t1;
--01创建序列
create sequence SEQ_SENDHDFS_ROUTE
minvalue 1
maxvalue 99999999999999999999
start with 1
increment by 1
cache 200;
--02创建job
declare waybill_basic_job number;
begin
dbms_job.submit(waybill_basic_job,'STP_TEMP_WAYBILL;',sysdate,'TRUNC(sysdate+1) +1/24');
commit;
end;
begin
dbms_job.remove(900);
commit;
end;
--03创建同义词
create synonym TM_CVYGAP_ROUTE for OMPROUTE.TM_CVYGAP_ROUTE;
commit;