oracle 舍位函数,如何协助 MySQL 实现 Oracle 高级分析函数

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概述

Oracle 支持一些独特的语法和函数，在移植到 MySQL 上时或多或少给程序员造成了困扰，下面我们针对 Oracle 的一些特殊用法举例并讲解如何用集算器来完成同样功能。这些方法当然也不限于针对 MySQL，对于所有其它数据库也能支持。

1、递归语句

a) select employee_id,first_name,last_name,manager_id

from hr.employees

start with employee_id=102

connect by prior employee_id = manager_idA

1=connect("orcl")

2=A1.query@x("select employee_id, first_name, last_name, manager_id from hr.employees")

3=A2.keys(EMPLOYEE_ID)

4=A2.select@1(EMPLOYEE_ID==102)

5=A2.switch(MANAGER_ID, A2)

6=A2.nodes(MANAGER_ID, A4)

7=(A4|A6).new(EMPLOYEE_ID, FIRST_NAME, LAST_NAME, MANAGER_ID.EMPLOYEE_ID:MANAGER_ID)

(1) A3 设置序表 A2 的键

(2) A4 选取起始雇员

(3) A5 将 A2 中 MANAGER_ID 值转换成记录，以便递归

(4) A6 获取起始雇员的所有子节点

b) select employee_id, first_name,last_name,manager_id

from hr.employees

start with employee_id=104

connect by prior manager_id = employee_idA

1=connect("orcl")

2=A1.query@x("select employee_id, first_name, last_name, manager_id from hr.employees")

3=A2.keys(EMPLOYEE_ID)

4=A2.switch(MANAGER_ID, A2)

5=A2.select@1(EMPLOYEE_ID==104)

6=A5.prior(MANAGER_ID)

7=A6.new(EMPLOYEE_ID, FIRST_NAME, LAST_NAME, MANAGER_ID.EMPLOYEE_ID:MANAGER_ID)

(1) A6 获取起始雇员的所有父节点

c) select employee_id,last_name,manager_id,sys_connect_by_path(last_name,'/') path from hr.employees

start with employee_id=102

connect by prior employee_id = manager_idA

1=connect("orcl")

2=A1.query@x("select employee_id, last_name, manager_id,null path from hr.employees")

3=A2.keys(EMPLOYEE_ID)

4=A2.select@1(EMPLOYEE_ID==102)

5=A2.switch(MANAGER_ID, A2)

6=A2.nodes(MANAGER_ID, A4)

7=A4|A6

8=A7.run(PATH=if(EMPLOYEE_ID==102, "/"+LAST_NAME, MANAGER_ID.PATH+"/"+LAST_NAME))

9=A7.new(EMPLOYEE_ID, LAST_NAME, MANAGER_ID.EMPLOYEE_ID:MANAGER_ID, PATH)

(1) 由于 A7 中每条记录的父节点都在本节点之前，故 A8 可以从前往后对 A7 中每条记录依次修改 PATH 值

2、嵌套聚集函数

select avg(max(salary)) avg_max, avg(min(salary)) avg_min

from hr.employees

group by department_idA

1=connect("orcl")

2=A1.query@x("select * from hr.employees")

3=A2.groups(DEPARTMENT_ID;max(SALARY):m1, min(SALARY):m2)

4=A3.group(;~.avg(m1):avg_max,~.avg(m2):avg_min)

(1) A2 中 A1.query 也可以改用 A1.cursor

3、聚集分析函数 FIRST 和 LAST

SELECT department_id,

MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY commission_pct) worst,

MAX(salary) KEEP (DENSE_RANK LAST ORDER BY commission_pct) best

FROM hr.employees

GROUP BY department_id

ORDER BY department_idA

1=connect("orcl")

2=A1.query@x("select * from hr.employees order by department_id,commission_pct")

3=A2.group@o(DEPARTMENT_ID)

4=A3.new(DEPARTMENT_ID,~.minp@a(ifn(COMMISSION_PCT,2)).min(SALARY):worst, ~.maxp@a(ifn(COMMISSION_PCT,2)).max(SALARY):best)

5=A4.sort(ifn(DEPARTMENT_ID,power(2,32)))

(1) A2 已按 DEPARTMENT_ID 排序，则 A3 分组时可采用 group@o

(2) FIRST/LAST 取排序的后第一组 / 最后一组，而 Oracle 排序时 null 排在最后，所以 LAST 会取到的最后一组就是 null 值所在组。maxp/minp 求具有最大值 / 最小值的所有行时排除了 null，所以在 A4 是用 ifn(COMMISSION_PCT,2) 保证 null 值时最大

(3) A5 中，DEPARTMENT_ID=null 时采用采用比所有 DEPARTMENT_ID 都大的 power(2,32) 来保证这一行排在最后

如果数据量大，还可以采用游标方式。A

1=connect("orcl")

2=A1.cursor@x("select * from hr.employees")

3=A2.groups(DEPARTMENT_ID; min([if(COMMISSION_PCT,2),SALARY]):m1, max([if(COMMISSION_PCT,2), SALARY]):m2)

4=A3.new(DEPARTMENT_ID, m1(2):worst, m2(2):best)

5=A4.sort(ifn(DEPARTMENT_ID,power(2,32)))

(1) A3 中，min([if(COMMISSION_PCT,2), SALARY]) 求出 COMMISSION_PCT 最小时的 SALARY 最小值，即 COMMISSION_PCT 排名第一时 SALARY 最小值，max 类似

4、占比函数 ratio_to_report

a) SELECT last_name, salary, RATIO_TO_REPORT(salary) OVER () AS rr

FROM hr.employees

WHERE job_id = 'PU_CLERK'

ORDER BY last_nameA

1=connect("orcl")

2=A1.query@x("select last_name,salary from hr.employees where job_id='PU_CLERK'order by last_name")

3=A2.sum(SALARY)

4=A2.new(LAST_NAME, SALARY, SALARY/A3:RR)

b) SELECT department_id,last_name, salary, RATIO_TO_REPORT(salary) OVER (partition by department_id) AS rr

FROM hr.employees

WHERE department_id in (20,60)

ORDER BY department_id,last_nameA

1=connect("orcl")

2=A1.query@x("select department_id,last_name,salary from hr.employees where department_id in (20,60) order by department_id,last_name")

3=A2.groups@o(DEPARTMENT_ID;sum(SALARY):sum)

4=A2.switch(DEPARTMENT_ID, A3)

5=A2.new(DEPARTMENT_ID.DEPARTMENT_ID:DEPARTMENT_ID, LAST_NAME, SALARY, SALARY/DEPARTMENT_ID.sum:RR)

(1) A2 中已按 DEPARTMENT_ID 排序，则 A3 可用 groups@o 分组聚集

5、多重分组

SELECT department_id, job_id, sum(salary) total

FROM hr.employees

WHERE department_id in (30, 50)

GROUP BY grouping sets((department_id, job_id), department_id)A

1=connect("orcl")

2=A1.query@x("select department_id, job_id, salary from hr.employees where department_Id in (30,50) order by department_id, job_id")

3=A2.groups@o(DEPARTMENT_ID, JOB_ID; sum(SALARY):TOTAL)

4=A3.group@o(DEPARTMENT_ID, null:JOB_ID; ~.sum(TOTAL):TOTAL)

5=[A3,A4].merge(DEPARTMENT_ID, ifn(JOB_ID,fill("z", 10)))

(1) 因为 A3 和 A4 均对 DEPARTMENT_ID 有序，故 A5 可 merge，ifn(JOB_ID,fill("z",10))) 用来保证 JOB_ID 为 null 排在后面

也可以采用游标方式。A

1=connect("orcl")

2=A1.cursor@x("select department_id,job_id,sum(salary) total from hr.employees where department_id in (30,50) group by department_id, job_id order by department_id,job_id")

3=A2.group(DEPARTMENT_ID)

4=A3.(~.insert(0, ~.groups@o(DEPARTMENT_ID, null:JOB_ID;sum(TOTAL):TOTAL)))

5=A4.fetch()

6=A5.conj()

(1) A3 中 A2.group 要求 A2 对 DEPARTMENT_ID 有序

(2) A4 对 A3 每一组求和并将结果插入此组末尾

还可以采用管道方式。A

1=connect("orcl")

2=A1.cursor@x("select department_id, job_id, salary from hr.employees where department_Id in (30,50) order by department_id, job_id")

3=channel().group@o(DEPARTMENT_ID, JOB_ID; sum(SALARY):TOTAL)

4>A2.push(A3)

5=channel().groups@o(DEPARTMENT_ID, null:JOB_ID; sum(TOTAL):TOTAL)