行业资讯

Mysql似oracle分析函数sum over的实现方法是什么

发布时间：2021-11-18 17:10:00 作者：iii
来源：亿速云 阅读：168

这篇文章主要介绍“Mysql似oracle分析函数sum over的实现方法是什么”，在日常操作中，相信很多人在Mysql似oracle分析函数sum over的实现方法是什么问题上存在疑惑，小编查阅了各式资料，整理出简单好用的操作方法，希望对大家解答”Mysql似oracle分析函数sum over的实现方法是什么”的疑惑有所帮助！接下来，请跟着小编一起来学习吧！

先看oracle怎么实现的

  1. select deptno,ename,sal,sum(sal) over(order by ename) from emp; --姓名排序连续求和

  2. select deptno,ename,sal,sum(sal) over(order by deptno) from emp; --所有部们排序连续求和

  3. select deptno,ename,sal,sum(sal) over(partition by deptno) from emp; ---各个部门的总和

  4. select deptno,ename,sal,sum(sal) over(partition by deptno order by ename) from emp; ---各个部门之间连续求和

  5. select deptno,ename,sal,sum(sal) over(order by deptno,ename) from emp;



  8. select deptno,ename,sal,

  9.       sum(sal) over (partition by deptno order by ename) 部门连续求和,--各部门的薪水"连续"求和

  10.       sum(sal) over (partition by deptno) 部门总和, -- 部门统计的总和，同一部门总和不变

  11.       100*round(sal/sum(sal) over (partition by deptno),4) "部门份额(%)",

  12.       sum(sal) over (order by deptno, ename) 连续求和, --所有部门的薪水"连续"求和

  13.       sum(sal) over () 总和, -- 此处sum(sal) over () 等同于sum(sal)，所有员工的薪水总和

  14.      100*round(sal/sum(sal) over (),4) "总份额(%)"

  15.      from emp



mysql的实现

  1. 如下：

  2. SELECT a.id,a.user_id,a.borrow_id, a.repayment_money,
    (SELECT SUM(repayment_money) FROM rb_repayment_period WHERE id<=a.id) "累加和",

  3. (SELECT AVG(repayment_money) FROM rb_repayment_period WHERE id<=a.id) "平均值" ,
    (SELECT SUM(repayment_money) FROM rb_repayment_period  WHERE borrow_id=a.borrow_id GROUP BY borrow_id) "每组和",
    (SELECT SUM(repayment_money) FROM rb_repayment_period) "全部和",
    (SELECT SUM(repayment_money) FROM rb_repayment_period  WHERE id<=a.id  GROUP BY borrow_id HAVING  borrow_id=a.`borrow_id` ) "每组累加和"    
    FROM rb_repayment_period a;

结果
Mysql似oracle分析函数sum over的实现方法是什么

原数据

  1. sql：

  2. CREATE TABLE `rb_repayment_period` (
      `id` int(11) NOT NULL AUTO_INCREMENT COMMENT '主键',
      `borrow_id` int(11) DEFAULT '0' COMMENT '标的id',
      `user_id` int(11) DEFAULT '0' COMMENT '借款人id',
      `repayment_money` decimal(20,6) DEFAULT '0.000000' COMMENT '本次还款金额',
      `capital_money` decimal(20,6) DEFAULT '0.000000' COMMENT '本金',
      `expect_money` decimal(20,6) DEFAULT '0.000000' COMMENT '预期收益',
      `exceed_money` decimal(20,6) DEFAULT '0.000000' COMMENT '超额收益',
      `actual_rate` decimal(20,6) DEFAULT '0.000000' COMMENT '实际收益率',
      `third_company_money` decimal(20,6) DEFAULT '0.000000' COMMENT '第三方公司收益',
      `load_money` decimal(20,6) DEFAULT '0.000000' COMMENT '借款人利益',
      `repayment_time` int(3) DEFAULT '0' COMMENT '还款次数',
      `repayment_stage` int(3) DEFAULT '0' COMMENT '当前还款的阶段',
      `playform_money` decimal(20,6) DEFAULT '0.000000' COMMENT '平台收益',
      `add_datetime` timestamp NOT NULL DEFAULT '2016-04-24 03:49:30' COMMENT '操作时间',
      `memo_id_first` int(11) DEFAULT '0' COMMENT '备用id',
      `memo_dec_first` decimal(20,6) DEFAULT '0.000000' COMMENT '备用dec',
      `memo_str_first` varchar(500) DEFAULT NULL COMMENT '备用str1',
      `memo_str_second` varchar(500) DEFAULT NULL COMMENT '备用str2',
      `memo_date_first` timestamp NULL DEFAULT '2016-04-24 03:49:30' COMMENT '备用时间1',
      `memo_date_second` timestamp NULL DEFAULT '2016-04-24 03:49:30' COMMENT '备用时间2',
      `total_repay_money` decimal(20,6) DEFAULT '0.000000' COMMENT '累计还款总额',
      `repay_type` int(3) DEFAULT '0' COMMENT '还款类型',
      `left_capital_money` decimal(20,6) DEFAULT '0.000000' COMMENT '剩余本金',
      `left_expect_money` decimal(20,6) DEFAULT '0.000000' COMMENT '剩余收益',
      `left_money` decimal(20,6) DEFAULT '0.000000' COMMENT '剩余留用',
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB AUTO_INCREMENT=60 DEFAULT CHARSET=utf8;
    /*!40101 SET character_set_client = @saved_cs_client */;


    --
    -- Dumping data for table `rb_repayment_period`
    --


    LOCK TABLES `rb_repayment_period` WRITE;
    /*!40000 ALTER TABLE `rb_repayment_period` DISABLE KEYS */;
    INSERT INTO `rb_repayment_period` VALUES (26,160,188,1000.000000,1000.000000,0.000000,0.000000,0.000000,0.000000,0.000000,1,2,0.0000
    00,'2016-04-24 07:43:38',0,0.000000,NULL,NULL,'2016-04-24 03:49:30','2016-04-24 03:49:30',0.000000,0,0.000000,0.000000,0.000000),(27
    ,160,188,100.000000,0.000000,100.000000,0.000000,0.000000,0.000000,0.000000,2,2,0.000000,'2016-04-24 07:45:26',0,0.000000,NULL,NULL,
    '2016-04-24 03:49:30','2016-04-24 03:49:30',0.000000,0,0.000000,0.000000,0.000000),(30,160,188,1000.000000,0.000000,87.500000,11.250
    000,0.000000,11.250000,890.000000,3,4,0.000000,'2016-04-24 08:09:11',0,0.000000,NULL,NULL,'2016-04-24 03:49:30','2016-04-24 03:49:30
    ',0.000000,0,0.000000,0.000000,0.000000),(42,163,187,4400.000000,2000.000000,375.000000,0.000000,0.000000,0.000000,2025.000000,1,3,0
    .000000,'2016-04-25 07:33:59',0,0.000000,NULL,NULL,'2016-04-25 07:33:59','2016-04-25 07:33:59',0.000000,0,0.000000,0.000000,0.000000
    ),(47,172,187,10000.000000,2000.000000,375.000000,12.500000,0.000000,12.500000,7600.000000,1,4,0.000000,'2016-04-26 02:48:05',0,0.00
    0000,NULL,NULL,'2016-04-26 02:48:05','2016-04-26 02:48:05',0.000000,0,0.000000,0.000000,0.000000),(48,174,187,10000.000000,2000.0000
    00,375.000000,12.500000,0.000000,12.500000,7600.000000,1,4,0.000000,'2016-04-26 03:23:41',0,0.000000,NULL,NULL,'2016-04-26 03:23:41'
    ,'2016-04-26 03:23:41',0.000000,0,0.000000,0.000000,0.000000),(49,157,187,3000.000000,1000.000000,120.000000,0.000000,0.000000,0.000
    000,1880.000000,1,3,0.000000,'2016-04-26 03:58:56',0,0.000000,NULL,NULL,'2016-04-26 03:58:56','2016-04-26 03:58:56',3000.000000,2,0.
    000000,0.000000,0.000000),(50,175,187,10000.000000,2000.000000,375.000000,12.500000,0.000000,12.500000,7600.000000,1,4,0.000000,'201
    6-04-26 05:29:48',0,0.000000,NULL,NULL,'2016-04-26 05:29:48','2016-04-26 05:29:48',10000.000000,2,0.000000,0.000000,0.000000),(54,17
    7,187,2000.000000,2000.000000,0.000000,0.000000,0.000000,0.000000,0.000000,1,2,0.000000,'2016-04-27 01:59:35',0,0.000000,NULL,NULL,'
    2016-04-27 01:59:35','2016-04-27 01:59:35',2000.000000,1,0.000000,375.000000,0.000000),(55,177,187,4000.000000,0.000000,375.000000,0
    .000000,360.000000,0.000000,3625.000000,2,3,0.000000,'2016-04-27 02:01:43',0,0.000000,NULL,NULL,'2016-04-27 02:01:43','2016-04-27 02
    :01:43',6000.000000,2,0.000000,0.000000,0.000000),(56,178,187,2100.000000,2000.000000,100.000000,0.000000,0.000000,0.000000,0.000000
    ,1,2,0.000000,'2016-04-27 03:43:43',0,0.000000,NULL,NULL,'2016-04-27 03:43:43','2016-04-27 03:43:43',2100.000000,1,0.000000,275.0000
    00,0.000000),(57,178,187,3000.000000,0.000000,275.000000,0.000000,378.000000,0.000000,2725.000000,2,3,0.000000,'2016-04-27 07:07:34'
    ,0,0.000000,NULL,NULL,'2016-04-27 07:07:34','2016-04-27 07:07:34',5100.000000,2,0.000000,0.000000,0.000000),(58,181,187,1000.000000,
    1000.000000,0.000000,0.000000,0.000000,0.000000,0.000000,1,1,0.000000,'2016-04-27 07:15:58',0,0.000000,NULL,NULL,'2016-04-27 07:15:5
    8','2016-04-27 07:15:58',1000.000000,1,1000.000000,375.000000,0.000000),(59,181,187,500.000000,500.000000,0.000000,0.000000,180.0000
    00,0.000000,0.000000,2,1,0.000000,'2016-04-27 07:26:34',0,0.000000,NULL,NULL,'2016-04-27 07:26:34','2016-04-27 07:26:34',1500.000000
    ,1,500.000000,375.000000,0.000000);

rownum的实现

  1. 环境：

  2. mysql> show create table tbl\G;
    *************************** 1. row ***************************
           Table: tbl
    Create Table: CREATE TABLE `tbl` (
      `id` int(11) NOT NULL,
      `col` int(11) DEFAULT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8

  3. mysql> insert into tbl values (1,26),(2,46),(3,35),(4,68),(5,93),(6,92);

  4. mysql> select * from tbl
        -> ;
    +----+------+
    | id | col  |
    +----+------+
    |  1 |   26 |
    |  2 |   46 |
    |  3 |   35 |
    |  4 |   68 |
    |  5 |   93 |
    |  6 |   92 |
    +----+------+
    6 rows in set (0.00 sec)



  7. 实现一：





  8. 实现二：解决重复bug（先建立一张数字表Nums(a int) 插入1-100即可）


    第二步：
    MySQL [interface_hd_com]> select Nums.a+c.rownum as rank ,col from (select a.col,COUNT(*) as count,( select count(*) from testtt b where b.col<a.col) as rownum from testtt a group by a.col) c,Nums where Nums.a<=count order by col;
    +------+------+
    | rank | col  |
    +------+------+
    |    1 |   26 |
    |    2 |   35 |
    |    3 |   35 |
    |    4 |   46 |
    |    5 |   46 |
    |    6 |   68 |
    |    7 |   68 |
    |    8 |   92 |
    |    9 |   92 |
    |   10 |   93 |
    |   11 |   93 |
    +------+------+
    11 rows in set (0.01 sec)






    连续区间的实现（求连续id区间）

    第二步：计算一下与标示的差值（如果是连续的，那么差值一样）
    mysql> SELECT id,alias1,(id-alias1) AS diff FROM (SELECT id,@id:=@id+1  AS alias1 FROM tbl,(SELECT @id:=0) AS id) b;
    +----+--------+------+
    | id | alias1 | diff |
    +----+--------+------+
    | 11 |      1 |   10 |
    | 12 |      2 |   10 |
    | 13 |      3 |   10 |
    | 14 |      4 |   10 |
    | 15 |      5 |   10 |
    | 16 |      6 |   10 |
    | 18 |      7 |   11 |
    | 19 |      8 |   11 |
    +----+--------+------+
    8 rows in set (0.00 sec)


    第三步：根据差值分组找出最大最小即可
    mysql> SELECT MIN(id) start_pos,MAX(id) end_pos
        -> FROM
        -> (SELECT id,alias1,(id-alias1) AS diff FROM (SELECT id,@id:=@id+1  AS alias1 FROM tbl,(SELECT @id:=0) AS id) b)
        -> AS c
        -> GROUP BY diff;
    +-----------+---------+
    | start_pos | end_pos |
    +-----------+---------+
    |        11 |      16 |
    |        18 |      19 |
    +-----------+---------+
    2 rows in set (0.00 sec)



    实验：求tel相同的连续段

    按照上面的思路求得
    MySQL [interface_hd_com]> SELECT MIN(id) start_pos,MAX(id) end_pos,tel FROM (SELECT id,alias1,(id-alias1) AS diff,tel FROM (SELECT id,@id:=@id+1  AS alias1,tel FROM testtab,(SELECT @id:=0) AS id) b) as c GROUP BY diff,tel order by tel desc;
    +-----------+---------+--------+
    | start_pos | end_pos | tel    |
    +-----------+---------+--------+
    |         3 |       7 | 187164 |
    |         1 |       8 | 187163 |
    |         9 |       9 |  19999 |
    +-----------+---------+--------+   ---这样是有bug的

    发现这样是不行的，因为id是连续的，所以同一个tel的diff是相同的，但其实中间隔着别的tel
    解决办法：分两次求在合并

    union 一下

    1. MySQL [interface_hd_com]> SELECT MIN(id) start_pos,MAX(id) end_pos,tel FROM (SELECT id,alias1,(id-alias1) AS diff,tel FROM (SELECT id,@id:=@id+1  AS alias1,tel FROM testtab,(SELECT @id:=0) AS id where tel in (SELECT distinct(tel) from testtab where tel<>187164)) b) as c GROUP BY diff,tel order by tel desc;

    2. +-----------+---------+--------+

    3. | start_pos | end_pos | tel |

    4. +-----------+---------+--------+

    5. | 1 | 2 | 187163 |

    6. | 5 | 6 | 187163 |

    7. | 8 | 8 | 187163 |

    8. | 9 | 9 | 19999 |

    9. +-----------+---------+--------+

    10. 4 rows in set (0.00 sec)


    12. MySQL [interface_hd_com]> SELECT MIN(id) start_pos,MAX(id) end_pos,tel FROM (SELECT id,alias1,(id-alias1) AS diff,tel FROM (SELECT id,@id:=@id+1  AS alias1,tel FROM testtab,(SELECT @id:=0) AS id where tel in (187164)) b) as c GROUP BY diff,tel order by tel desc;

    13. +-----------+---------+--------+

    14. | start_pos | end_pos | tel |

    15. +-----------+---------+--------+

    16. | 3 | 4 | 187164 |

    17. | 7 | 7 | 187164 |

    18. +-----------+---------+--------+

    19. 2 rows in set (0.00 sec)

    20. MySQL [interface_hd_com]> select * from testtab;

    21. +------+--------+

    22. | id | tel |

    23. +------+--------+

    24. | 1 | 187163 |

    25. | 2 | 187163 |

    26. | 3 | 187164 |

    27. | 4 | 187164 |

    28. | 5 | 187163 |

    29. | 6 | 187163 |

    30. | 7 | 187164 |

    31. | 8 | 187163 |

    32. | 9 | 19999 |

    33. +------+--------+

    34. 9 rows in set (0.00 sec)

    35. 第一步：标示

    36. mysql> SELECT id,@id:=@id+1  AS alias1 FROM tbl,(SELECT @id:=0) AS id;
      +----+--------+
      | id | alias1 |
      +----+--------+
      | 11 |      1 |
      | 12 |      2 |
      | 13 |      3 |
      | 14 |      4 |
      | 15 |      5 |
      | 16 |      6 |
      | 18 |      7 |
      | 19 |      8 |
      +----+--------+
      8 rows in set (0.00 sec)

    37. 第一步求出个数

    38. MySQL [interface_hd_com]> select a.col,COUNT(*) as count,( select count(*) from testtt b where b.col<a.col) as rownum from testtt a group by a.col;

    39. +------+-------+--------+

    40. | col | count | rownum |

    41. +------+-------+--------+

    42. | 26 | 1 | 0 |

    43. | 35 | 2 | 1 |

    44. | 46 | 2 | 3 |

    45. | 68 | 2 | 5 |

    46. | 92 | 2 | 7 |

    47. | 93 | 2 | 9 |

    48. +------+-------+--------+

    49. 6 rows in set (0.00 sec)

    50. mysql> select id,a.col,( select count(*) from tbl b where b.col<=a.col) as rank from tbl a order by rank;
      +----+------+------+
      | id | col  | rank |
      +----+------+------+
      |  1 |   26 |    1 |
      |  3 |   35 |    2 |
      |  2 |   46 |    3 |
      |  4 |   68 |    4 |
      |  6 |   92 |    5 |
      |  5 |   93 |    6 |
      +----+------+------+
      6 rows in set (0.00 sec)

    51. 瑕疵：当有重复的数据时就有bug了

    52. mysql> select id,a.col,(select count(*) from tbl b where b.col<=a.col ) as rank from tbl a order by rank;
      +----+------+------+
      | id | col  | rank |
      +----+------+------+
      |  1 |   26 |    2 |
      |  9 |   26 |    2 |
      |  3 |   35 |    4 |
      |  8 |   35 |    4 |
      |  2 |   46 |    5 |
      |  4 |   68 |    6 |
      |  6 |   92 |    7 |
      |  5 |   93 |    8 |
      +----+------+------+
      8 rows in set (0.00 sec)



到此，关于“Mysql似oracle分析函数sum over的实现方法是什么”的学习就结束了，希望能够解决大家的疑惑。理论与实践的搭配能更好的帮助大家学习，快去试试吧！若想继续学习更多相关知识，请继续关注亿速云网站，小编会继续努力为大家带来更多实用的文章！

推荐阅读：
  1. oracle菜鸟学习之 分析函数-排序
  2. oracle 函数之分析函数

免责声明：本站发布的内容（图片、视频和文字）以原创、转载和分享为主，文章观点不代表本网站立场，如果涉及侵权请联系站长邮箱：is@yisu.com进行举报，并提供相关证据，一经查实，将立刻删除涉嫌侵权内容。

mysql oracle

上一篇：Ceph如何实现集群搭建

下一篇：Docker Registry v2如何搭建

相关阅读

您好，登录后才能下订单哦！

密码登录
登录注册
获取短信验证码
其他方式登录
点击 登录注册 即表示同意《亿速云用户服务条款》
产品服务
地区划分
帮助支持
关于我们
行业资讯-文章归档 问答-问答归档
广州亿速云计算有限公司

7*24小时在线电话：400-100-2938

7*24小时在线QQ：800811969

Copyright © Yisu Cloud Ltd. All Rights Reserved. 2018 版权所有
粤ICP备17096448号-1 粤公网安备 44010402001142号