mysql实现row_number的方法

1、利用内联关系取出多组条数据最新的数据(即oracle的row_number概念):
-- 筛选出的状态 inner join筛选出来的状态值

select t.flow_id,t.job_id,t.`status`,t.start_time

from (

SELECT flow_id,job_id,status,start_time,max(FROM_UNIXTIME(start_time/1000)) as max_start_time

from execution_jobs 

-- where (flow_id='end_it_sda' or flow_id='end_paasjszb_sda')

where (flow_id='end_paasjszb_sda') 

and start_time>1544976295990 

-- and job_id like '%sda%'

group by job_id,status

) t,(

--内联用例

SELECT flow_id,job_id,`status`,start_time,max(FROM_UNIXTIME(start_time/1000)) as max_start_time

from execution_jobs

-- where (flow_id='end_it_sda' or flow_id='end_paasjszb_sda')

where (flow_id='end_paasjszb_sda')

and start_time>1544976295990

-- and job_id='it_sda01_jiraissue'

group by job_id

) t1

where t1.max_start_time=t.max_start_time and t1.job_id=t.job_id

2
、按照分区来查mysql数据: select etl_tx_dt ,count(*) from zcfw_sda.sda01_core_asset_bad_out_info group by etl_tx_dt ; 3、启动django项目命令:django-admin startproject HelloWorld 4、通过分组确定唯一标识: select count(*),core_lend_request_id,lend_contract_code,loan_time from datawindow_dw.finance_report_defferred_income_detail_2 where etl_tx_dt=20181208 group by core_lend_request_id,lend_contract_code,loan_time having count(*)>1; 5、分组按照日期排序: select etl_tx_dt ,count(*) from hhpt_sda.sda01_btows_batch_mobile_20181210 group by etl_tx_dt order by etl_tx_dt desc; 6、数据库操作: (1)模糊匹配查找出表execution_jobs下flow_id有sda名字的 SELECT distinct flow_id from execution_jobs where execution_jobs.flow_id like 'sda%'; (2)模糊匹配查找出表projects下有sda_pro名字并且active=1的所有信息 select * from projects p where p.active = 1 and p.name like '%sda_pro'; (3)从表execution_jobs找出10个时间戳为秒级别的从大到小的排序 select FROM_UNIXTIME(el.start_time/1000),el.* from execution_jobs el order by FROM_UNIXTIME(el.start_time/1000) desc limit 104)从表execution_jobs找出时间戳大于2018-12-12的所有信息 select FROM_UNIXTIME(el.start_time/1000),el.* from execution_jobs el where FROM_UNIXTIME(el.start_time/1000) > '2018-12-12'5)统计表execution_jobs中时间戳大于2018-12-12的条数 select count(*) from execution_jobs el where FROM_UNIXTIME(el.start_time/1000) > '2018-12-12'; (6)从表projects中找出时间戳大于2018-12-12,active=1和name中有sda_pro的任务列表,其中匹配条件为表project的id等于表execution_jobs的id select p.`name`,el.job_id from projects p,execution_jobs el where p.active = 1 and p.name like '%sda_pro' and p.id = el.project_id and FROM_UNIXTIME(el.start_time/1000) > '2018-12-12' group by p.`name`,el.job_id order by p.`name`,el.job_id; (7)从表projects中找出时间戳大于2018-12-12,active=1和name中有sda_pro,并且execution_jobs表中job_id也有sda_pro的任务列表,其中匹配条件
为表project的id等于表execution_jobs的id
select p.`name`,el.job_id from projects p,execution_jobs el where p.active = 1 and p.name like '%sda_pro' and p.id = el.project_id and FROM_UNIXTIME(el.start_time/1000) > '2018-12-12' and el.job_id like '%sda%' group by p.`name`,el.job_id order by p.`name`,el.job_id; 7、时间戳与日期的相互转换: -- 时间戳转换成时间格式 select FROM_UNIXTIME(1544991635000/1000) -- 日期转换成时间戳 SELECT UNIX_TIMESTAMP('2018-12-22 00:00:00') *1000

 

posted @ 2018-12-17 23:51  极客浩浩  阅读(4398)  评论(0编辑  收藏  举报