软件测试面试题夫唯seo

PGSQL基本使用

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日期转换

格式	说明
YYYY	年
MM	月
DD	日
hh24	24小时制
mi	分钟
ss	秒

-- 日期字符串转指定日期字符串
-- 20250101123000 转为 2025-01-01 12:30:00
select to_char(to_timestamp('20250101123000','YYYYMMDDhh24miss'),'YYYY-MM-DD hh24:mi:ss') as t;

长度不够补数

不足前面补数

-- 得到字符串 '09300000'
select lpad(cast(9300000 as text),8,'0') as t;

不足前面补数

-- 得到字符串 '15300000'
select lpad(cast(153000 as text),8,'0') as t;

获取上下行

获取上一行

-- 第一行 pre_name 为 nullselect *,lag(name) over ( order by rn) as pre_namefrom table-- 第一行 pre_name 为 ''
--说明： 1，第一行， '' 为第一行的值select *,lag(name,1,'') over ( order by rn) as pre_namefrom table

获取下一行

-- 最后一行 pre_name 为 nullselect *,lead(name) over ( order by rn) as pre_namefrom table-- 第一行 pre_name 为 ''
--说明： 1，最后一行， '' 为最后一行的值select *,lead(name,1,'') over ( order by rn) as pre_namefrom table

取连续的开始和结束的值

同一个账户，订单编号需要去除，5,9 。计算订单编号连续

-- 基础数据
with base as (select  1 as order_num, 'A01' as acct_idunion all select  2 as order_num, 'A01' as acct_idunion all select  3 as order_num, 'A02' as acct_idunion all select  4 as order_num, 'A02' as acct_idunion all select  5 as order_num, 'A02' as acct_idunion all select  6 as order_num, 'A02' as acct_idunion all select  7 as order_num, 'A01' as acct_idunion all select  8 as order_num, 'A01' as acct_idunion all select  9 as order_num, 'A01' as acct_idunion all select  10 as order_num, 'A01' as acct_idunion all select  11 as order_num, 'A01' as acct_id),-- 根据acct_id分组，根据order_num排序
base_order as (select *, row_number() over (partition by acct_id order by order_num) as rn from base
),
-- 获取上下行的rn
-- lag(rn,1,null) 表示第一行的 rn值为空
-- lead(rn,1,null) 表示最后一行的 rn值为空
base_rn as (
select *,lag(rn,1,null) over (partition by acct_id order by order_num) as pre_rn,lead(rn,1,null) over (partition by acct_id order by order_num) as next_rn
from base_order
where order_num != 5 and order_num != 9
),
-- 计算acct_id连续的 order_num
base_link as (
select *,case when pre_rn is null or pre_rn + 1 != rn then  order_num else null end as start_order_num, /**连续的开始*/case when next_rn is null or next_rn -1 != rn then order_num else null end as end_order_num, /**连续的结束*/sum(case when pre_rn is null or pre_rn + 1 != rn then 1 else 0 end) over (partition by acct_id order by order_num) as link_cnt /**连续的分为一组*/
from base_rn
)
-- 获取连续的开始订单编号和对应的结束订单编号
select max(start_order_num) as start_order_num,max(end_order_num) as end_order_num, acct_id
from base_link
group by acct_id, link_cnt

得到结果

start_order_num	end_order_num	acct_id
1	2	A01
3	4	A02
6	6	A02
7	8	A01
10	11	A01