MySQL中怎么实现一个分析函数

本篇文章为大家展示了MySQL中怎么实现一个分析函数，内容简明扼要并且容易理解，绝对能使你眼前一亮，通过这篇文章的详细介绍希望你能有所收获。

成都创新互联公司2013年至今，是专业互联网技术服务公司，拥有项目成都做网站、网站建设网站策划，项目实施与项目整合能力。我们以让每一个梦想脱颖而出为使命，1280元渝中做网站,已为上家服务,为渝中各地企业和个人服务,联系电话:13518219792

1. 实现rownum

1 2	`SET` `@rn:=0;` `SELECT` `@rn:=@rn+1` `AS` `rownum ,e.*` `FROM` `emp e;`

MySQL中怎么实现一个分析函数

或者写成：

1	`SELECT` `@rn:=@rn + 1` `AS` `rownum ,e.*` `FROM` `emp e ,(SELECT` `@rn:=0) c`

2. 各种分析函数写法 (MySQL实现分析语句时可能遇到的各种计算问题)

2.1 sum() 实现

--SQL 执行顺序 ,FROM ,JOIN ,WHERE ,GROUP BY,HAVING ,ORDER BY ,SELECT,

在Oracle中分页语句的原始语句如下：

1	`SELECT` `E.*,` `SUM(SAL) OVER(PARTITION` `BY` `DEPTNO)` `AS` `COUNTOVER` `FROM` `EMP E;`

MySQL中怎么实现一个分析函数

SELECT E.*,

(SELECT SUMOVER

FROM (SELECT DEPTNO, SUM(SAL) AS SUMOVER

FROM EMP E1

GROUP BY DEPTNO) X

WHERE X.DEPTNO = E.DEPTNO) AS COUNTOVER

FROM EMP E

ORDER BY DEPTNO;

MySQL中怎么实现一个分析函数

Mysql中也是这么实现的：

SELECT E.*,

(SELECT SUMOVER

FROM (SELECT DEPTNO, SUM(SAL) AS SUMOVER

FROM emp E1

GROUP BY DEPTNO) X

WHERE X.DEPTNO = E.DEPTNO) AS COUNTOVER

FROM emp E

ORDER BY DEPTNO;

MySQL中怎么实现一个分析函数

2.2 row_number () 实现

1 2	`select` `e.* ,row_number() over(partition` `by` `deptno` `order` `by` `empno)` `as` `ROW_NUMBER` `from` `emp e;`

MySQL中怎么实现一个分析函数

我们的默认规则是在from后初始化变量。

SELECT E.*,

IF(@DEPTNO = DEPTNO, @RN := @RN + 1, @RN := 1) AS ROW_NUMBER,

@DEPTNO := DEPTNO AS VAR1

FROM EMP E, (SELECT @DEPTNO := '', @RN := 0) C

ORDER BY DEPTNO;

MySQL中怎么实现一个分析函数

SELECT E.*,

IF(@DEPTNO = DEPTNO, @RN := @RN + 1, @RN := 1) AS ROW_NUMBER,

@DEPTNO := DEPTNO AS VAR1

FROM EMP E, (SELECT @DEPTNO := '', @RN := 0) C

ORDER BY DEPTNO;

这个语句首先执行order by

2.3 求每个人员占他所在部门总工资的百分比

在Oracle中实现：

SELECT E.*,

TRUNC(SAL / SUM(SAL) OVER(PARTITION BY DEPTNO), 3) AS SALPERCENT

FROM EMP E

ORDER BY DEPTNO;

MySQL中怎么实现一个分析函数

SELECT E.*,

SAL / (SELECT SUMOVER

FROM (SELECT DEPTNO, SUM(SAL) AS SUMOVER

FROM emp E1

GROUP BY DEPTNO) X

WHERE X.DEPTNO = E.DEPTNO) AS SalPercent

FROM emp E

ORDER BY DEPTNO;

MySQL中怎么实现一个分析函数

2.4 求各个部门的总共工资

Oracle:

1	`SELECT` `e.* ,SUM(sal) OVER(PARTITION` `BY` `deptno)` `FROM` `emp e;`

MySQL中怎么实现一个分析函数

MySQL:

SELECT A.*,

ROUND(CAST(IF(@DEPTNO = DEPTNO, @MAX := @MAX, @MAX := SUMOVER) AS CHAR ),0) AS SUMOVER2,

@DEPTNO := DEPTNO AS VAR2

FROM (SELECT E.*,

IF(@DEPTNO = DEPTNO, @SUM := @SUM + SAL, @SUM := SAL) AS SUMOVER,

@DEPTNO := DEPTNO AS VAR1

FROM emp E, (SELECT @DEPTNO := '', @SUM := 0, @MAX := 0) C

ORDER BY DEPTNO) A

ORDER BY DEPTNO, SUMOVER DESC;

子查询的功能实现如下：

MySQL中怎么实现一个分析函数

下面是这个语句的结果

MySQL中怎么实现一个分析函数

2.5 拿部门第二的工资的人

首先我们拿第二名的，用Oracle很好实现，不论是第一还是第二。

SELECT *

FROM (SELECT E.*,

ROW_NUMBER() OVER(PARTITION BY DEPTNO ORDER BY SAL DESC) AS RN

FROM EMP E)

WHERE RN = 2;

MySQL中怎么实现一个分析函数

Mysql中第一这么实现：

在5.6版本，sql_mode非only_full_group_by的情况，我们可以使用如下方式实现

1	`set` `global` `sql_mode ='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';`

1	`SELECT` `` `FROM` `(SELECT` `e.` `FROM` `emp e` `ORDER` `BY` `deptno,sal ) a` `GROUP` `BY` `deptno;`

在SQL_MODE非only_full_group_by时，MySQL中的group by是只取第一行的，下面我们看取第二行的SQL。

SELECT *

FROM (SELECT E.*,

IF(@DEPTNO = DEPTNO, @RN := @RN + 1, @RN := 1) AS RN,

@DEPTNO := DEPTNO

FROM EMP E, (SELECT @RN := 0, @DEPTNO := 0) C

ORDER BY DEPTNO, SAL DESC) X

WHERE X.RN = 2;

MySQL中怎么实现一个分析函数

2.6 dense_rank()

dense_rank函数返回一个唯一的值，除非当碰到相同数据时，此时所有相同数据的排名都一样。

SELECT empno,

ename,

sal,

deptno,

rank() OVER(PARTITION BY deptno ORDER BY sal desc) as rank,

dense_rank() OVER(PARTITION BY deptno ORDER BY sal desc) as dense_rank

FROM emp e;

MySQL中怎么实现一个分析函数

MySQL的写法：

select

empno,ename,sal,deptno,

if(@deptno = deptno,if(@sal=sal,@rn:=@rn,@rn3:=@rn3+1),@rn:=1) as "RANK() OVER",

if(@sal =sal,@rn2:=@rn2 ,if(@deptno = deptno,@rn2:=@rn2+1,@rn2:=1)) as "DENSE_RANK() OVER",

if(@deptno = deptno,@rn:=@rn+1,@rn:=1) as "ROW_NUMBER() OVER"

, @deptno:=deptno,@sal:=sal

from

(select empno,ename,sal,deptno from emp a ,(select @rn:=1,@deptno:=0,@rn2:=0,@rn3:=0,@sal:=0,@i:=0) b order by deptno,sal desc) c;

MySQL中怎么实现一个分析函数

2.7 连续获得冠军的有哪些

--请写出一条SQL语句，查询出在此期间连续获得冠军的有哪些，其连续的年份的起止时间是多少，结果如下：

create table nba as

SELECT '公牛' AS TEAM, '1991' AS Y FROM DUAL UNION ALL

SELECT '公牛' AS TEAM, '1992' AS Y FROM DUAL UNION ALL

SELECT '公牛' AS TEAM, '1993' AS Y FROM DUAL UNION ALL

SELECT '活塞' AS TEAM, '1990' AS Y FROM DUAL UNION ALL

SELECT '火箭' AS TEAM, '1994' AS Y FROM DUAL UNION ALL

SELECT '火箭' AS TEAM, '1995' AS Y FROM DUAL UNION ALL

SELECT '公牛' AS TEAM, '1996' AS Y FROM DUAL UNION ALL

SELECT '公牛' AS TEAM, '1997' AS Y FROM DUAL UNION ALL

SELECT '公牛' AS TEAM, '1998' AS Y FROM DUAL UNION ALL

SELECT '马刺' AS TEAM, '1999' AS Y FROM DUAL UNION ALL

SELECT '湖人' AS TEAM, '2000' AS Y FROM DUAL UNION ALL

SELECT '湖人' AS TEAM, '2001' AS Y FROM DUAL UNION ALL

SELECT '湖人' AS TEAM, '2002' AS Y FROM DUAL UNION ALL

SELECT '马刺' AS TEAM, '2003' AS Y FROM DUAL UNION ALL

SELECT '活塞' AS TEAM, '2004' AS Y FROM DUAL UNION ALL

SELECT '马刺' AS TEAM, '2005' AS Y FROM DUAL UNION ALL

SELECT '热火' AS TEAM, '2006' AS Y FROM DUAL UNION ALL

SELECT '马刺' AS TEAM, '2007' AS Y FROM DUAL UNION ALL

SELECT '凯尔特人' AS TEAM, '2008' AS Y FROM DUAL UNION ALL

SELECT '湖人' AS TEAM, '2009' AS Y FROM DUAL UNION ALL

SELECT '湖人' AS TEAM, '2010' AS Y FROM DUAL;

Oracle实现：

SELECT TEAM, MIN(Y), MAX(Y)

FROM (SELECT E.*,

ROWNUM,

ROW_NUMBER() OVER(PARTITION BY TEAM ORDER BY Y) AS RN,

ROWNUM - ROW_NUMBER() OVER(PARTITION BY TEAM ORDER BY Y) AS DIFF

FROM NBA E

ORDER BY Y)

GROUP BY TEAM, DIFF

HAVING MIN(Y) != MAX(Y)

ORDER BY 2;

MySQL实现：

SELECT TEAM, MIN(Y), MAX(Y)

FROM (SELECT TEAM,

Y,

IF(@TEAM = TEAM, @RN := @RN + 1, @RN := 1) AS RWN,

@RN1 := @RN1 + 1 AS RN,

@TEAM := TEAM

FROM nba N, (SELECT @RN := 0, @TEAM := '', @RN1 := '') C) A

GROUP BY RN - RWN

HAVING MIN(Y) != MAX(Y)

ORDER BY 2

MySQL中怎么实现一个分析函数

| UDF插件

Userdefined Function，用户定义函数。我们知道，MySQL本身支持很多内建的函数，此外还可以通过创建存储方法来定义函数。UDF为用户提供了一种更高效的方式来创建函数。

UDF与普通函数类似，有参数，也有输出。分为两种类型：单次调用型和聚集函数。前者能够针对每一行数据进行处理，后者则用于处理Group By这样的情况。

UDF自定义函数，在MySQL basedir/include

[root@test12c include]# pwd

/usr/local/mysql/include

[root@test12c include]# cat rownum.c

#include

#if defined(MYSQL_SERVER)

#include /* To get strmov() */

#else

/* when compiled as standalone */

#include

#define strmov(a,b) stpcpy(a,b)

#endif

#include

/*

gcc -fPIC -Wall -I/usr/local/mysql/include -I. -shared rownum.c -o rownum.so

DROP FUNCTION IF EXISTS rownum;

CREATE FUNCTION rownum RETURNS INTEGER SONAME 'rownum.so';

*/

C_MODE_START;

my_bool rownum_init(UDF_INIT *initid, UDF_ARGS *args, char *message);

void rownum_deinit(UDF_INIT *initid);

chong rownum(UDF_INIT *initid, UDF_ARGS *args, char *is_null,char *error);

C_MODE_END;

/*

Simple example of how to get a sequences starting from the first argument

or 1 if no arguments have been given

*/

my_bool rownum_init(UDF_INIT *initid, UDF_ARGS *args, char *message)

{

if (args->arg_count > 1)

{

strmov(message,"This function takes none or 1 argument");

return 1;

}

if (args->arg_count)

args->arg_type[0]= INT_RESULT; /* Force argument to int */

if (!(initid->ptr=(char*) malloc(sizeof(chong))))

{

strmov(message,"Couldn't allocate memory");

return 1;

}

memset(initid->ptr, 0, sizeof(chong));

initid->const_item=0;

return 0;

}

void rownum_deinit(UDF_INIT *initid)

{

if (initid->ptr)

free(initid->ptr);

}

chong rownum(UDF_INIT *initid __attribute__((unused)), UDF_ARGS *args,char *is_null __attribute__((unused)),char *error __attribute__((unused)))

{

uchong val=0;

if (args->arg_count)

val= *((chong*) args->args[0]);

return ++*((chong*) initid->ptr) + val;

}

生成动态链接库

1	`gcc rownum.c -fPIC -shared -o ../lib/plugin/rownum.so`

MySQL中怎么实现一个分析函数

上述内容就是MySQL中怎么实现一个分析函数，你们学到知识或技能了吗？如果还想学到更多技能或者丰富自己的知识储备，欢迎关注创新互联行业资讯频道。

分享题目：MySQL中怎么实现一个分析函数
分享地址：http://dzwzjz.com/article/gohcis.html

大橙子网站建设，新征程启航

MySQL中怎么实现一个分析函数

1. 实现rownum

2. 各种分析函数写法 (MySQL实现分析语句时可能遇到的各种计算问题)

2.1 sum() 实现

2.2 row_number () 实现

2.3 求每个人员占他所在部门总工资的百分比

2.4 求各个部门的总共工资

2.5 拿部门第二的工资的人

2.6 dense_rank()

2.7 连续获得冠军的有哪些

| UDF插件

其他资讯