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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_id
A | |
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_id
A | |
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_id
A | |
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_id
A | |
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_id
A | |
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_name
A | |
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_name
A | |
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) |
6 | >A3.push(A5) |
7 | =A3.fetch() |
8 | for A2,1000 |
9 | =A3.result()|A5.result() |
10 | =A9.sort(DEPARTMENT_ID) |
(1) A3 创建管道，并附加分组求和
(2) A4 将 A2 中数据推送到 A3，注意此动作只有在 A2 中数据有实际取出行为才执行
(3) A5 创建管道，并附加分组求和
(4) A6 将 A3 结果推送到 A5，此处也可以直接将 A2 中数据推送到 A5，但会增加时间复杂度
(5) A7 保留 A3 的数据
(6) 循环读取 A2，每次只取 1000 条，减少内存占用
(7) A10 对 A3 和 A5 中数据排序，因为算法是稳定的，所以 JOB_ID 为 null 的排在后面
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