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Oracle concept II

2022-06-27 16:37:00 【_ Become popular_】

There are two kinds of people in the world

Those who are brave and responsible and those who look for support

The backer is better

          ----《Scent of a Woman》 

Catalog

One .pivot function

Two .unpivot function

3、 ... and .with as

Four .leading function

5、 ... and .connect by recursive

On the left

On the right

6、 ... and .keep function

explain

7、 ... and .over function

 1、rank()/dense_rank() over(partition by ...order by ...)

 2、min()/max() over(partition by ...)

 3、lead()/lag() over(partition by ... order by ...)

lead() It's after taking N Row data

lag() It's before taking N Row data

One .pivot function

Transfer line column .

I met a business in my work , A column of data is required （ This column usually has different types , For example, county , Subjects, etc ） Convert to multiple columns to display , It can also be said to be row to column .

quote ： Blog

Student transcript , The original data ：

select class_name, student_name, course_type, result, created_date
  from class_tmp_2;

Each student's two grades , There are two pieces of data , Because business requires , When showing it to the user , The user wants only one piece of data for each student , And show all the results , such as

 

In this case, the row specific column function is required PIVOT.

SELECT class_name, student_name,  Chinese language and literature ,  mathematics , created_date
  FROM (SELECT CLASS_NAME, STUDENT_NAME, COURSE_TYPE, RESULT, CREATED_DATE
          FROM CLASS_TMP_2) T
PIVOT(SUM(RESULT)
   FOR COURSE_TYPE IN(' Chinese language and literature ' AS  Chinese language and literature , ' mathematics ' AS  mathematics ));

Light gray sql And the original data above sql equally , Mainly look at the back PIVOT part .

sum(result): Sum of achievements （PIVOT The aggregate function is required ）

for course_type in (' Chinese language and literature ' as Chinese language and literature , ' mathematics ' as mathematics ): take course_type Column Field values are converted to column names , among , The field value is ' Chinese language and literature ', Convert to language column , The field value is ' mathematics ', Convert to mathematical column , The field values of these two columns , I.e. the one in front sum(result).

This example belongs to the same class , Therefore, you can also use the following sub query to realize 

select t1.class_name,
       t1.student_name,
       t1.result        Chinese language and literature ,
       t2.result        mathematics ,
       t1.created_date
  from (select a.class_name,
               a.student_name,
               a.course_type,
               a.result,
               a.created_date
          from class_tmp_2 a
         where a.course_type = ' Chinese language and literature ') t1,
       (select a.class_name,
               a.student_name,
               a.course_type,
               a.result,
               a.created_date
          from class_tmp_2 a
         where a.course_type = ' mathematics ') t2
 where t1.class_name = t2.class_name
   and t1.student_name = t2.student_name;

Two .unpivot function

Column turned .

The original data are as follows ：

The desired results are as follows

in other words , If there are multiple columns to be found in the database, they will be displayed in multiple rows according to the fact that they are all placed in the same column , Their corresponding data is put into another category , That is, below result, Use this function .

select class_name, student_name, course_type, result, created_date
  from class_tmp 
  unpivot(result for course_type in(chinese_result,math_result));

explain ：

Original data chinese_result Column sum math_result The column name of the column （ Light grey ）, Convert to new column course_type The field values of the , Indicates the class type .

Original data chinese_result Column sum math_result Column's field value , Convert to new column result The field values of the , It's a score .

3、 ... and .with as

with as The phrase , Also called subquery part , Is to define a sql fragment , This piece is in the whole big sql It is often used in sentences , Then its use is limited to --- Used multiple times , Business scenarios with a small amount of data , increase sql Readability , Reduce the number of meter scanning （ This one uses F5 You can directly see whether it is the base table or the temporary table in memory ）, Reduce code rewriting ;with as Is equivalent to a temporary table in memory , So he quickly , He has good performance , So the amount of data cannot be very large ;

grammar ：

 The main way of writing ：
with query1 AS
(select ...from ....where ..),
query2  AS
(select...from ...where..),
query3 AS
(select...from ...where..)
SELECT ...FROM query1,quer2,query3 
where ....;

such as ：

with
e as
 (select * from scott.emp),
d as
 (select * from scott.dept)
select * from e, d where e.deptno = d.deptno;

Another example is multiple union all：

with

sql1 as
 (select to_char(a) s_name from test_tempa),

sql2 as
 (select to_char(b) s_name
    from test_tempb
   where not exists (select s_name from sql1 where rownum = 1))

select * from sql1
union all
select * from sql2
union all .. .. ..

Four .leading function

/*+LEADING(TABLE)*/      /*+LEADING(table1,table2....)*/
Make the specified table the first table in the link order ,leading Prompt can be followed by multiple table names , It is used to represent in related table associations , Which table is used as the driving table . Use leading After the prompt , The optimizer will no longer consider from The order of the following table .

Simply put, it specifies which is the driver table , It is usually a small watch ; sometimes oracle The optimizer will specify the large table as the driving table , This increases the number of scans ;

For example, I met a business recently , Due to confidentiality requirements , Similar statements are as follows 

select a...
  from A a
  left join B b
    on a.id = b.id
 inner join C c
    on a.id = c.id
 inner join D d
    on a.id = d.id

id yes ABCD The primary keys of the four tables , So this means that there are indexes , So, after looking at the scanning plan, I found A And B All tables are full table scanning ,B surface cost Longer time consuming , The total time is 45 Seconds or so ,A The table data has 100 strip ,B Table has 30 Ten thousand ;

In view of the above phenomenon , The first optimization direction is to make B Table mandatory index , So the optimization is as follows ：

select /*+index(B)*/ a...  
  from A a
  left join B
    on a.id = b.id
 inner join C c
    on a.id = c.id
 inner join D d
    on a.id = d.id

Looking at the implementation plan, we find that B The table is indexed , So the speed suddenly reached 0.2 second .

Continue with the example above , In fact, the final optimization is like this ：

select /*+leading(A B) index(B)*/ a...
  from A a
  left join B
    on a.id = b.id
 inner join C c
    on a.id = c.id
 inner join D d
    on a.id = d.id

Be careful ： If ⼀ Tables have indexes ,⼀ Tables have no indexes ,ORACLE The table without index will be used as the driving table . If both tables have indexes , The outer surface is used as the driving table . If both are not indexed , Is also the appearance of the driving table . Consider using... At this time leading Function to specify according to the scan plan .

5、 ... and .connect by recursive

Basic grammar 

select t.*,level from table t [start with condition1]
    connect by [prior] id=parentid

This function is generally used to find data with parent-child relationship , Tables with hierarchical relationships ;

  • start with condition1 It is used to limit the data of the first layer , Or root node data ; Based on this part of data, find the second layer data , Then use the second layer data to find the third layer data, and so on .

and connect by prior id=parentid perhaps connect by id=prior parentid What's the difference ？

Let's give you an example ：

Make a data , Take Naoki hamazawa as an example ：

-- Create table
create table BANK
(
  id         NUMBER(8) not null,
  funcid     NUMBER(8),
  parentid   NUMBER(8),
  funcname   VARCHAR2(20),
  parentname VARCHAR2(20)
);

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1001, 0, ' President Nakano Watanabe ', null);

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1002, 1001, ' Sanli, vice president ', ' President Nakano Watanabe ');

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1003, 1002, ' Executive of Ohata ', ' Sanli, vice president ');

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1005, 1003, ' Minister izoshan ', ' Executive of Ohata ');

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1006, 1005, ' Hanzawa Naoki ', ' Minister izoshan ');

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1007, 1003, ' Duzhen Li Ren ', ' Executive of Ohata ');

insert into BANK (FUNCID, PARENTID, FUNCNAME, PARENTNAME)
values (1004, 1002, ' The chronicle is standing ', ' Sanli, vice president ');

commit;

The query is shown in the figure , The hierarchical relationship is ： governor - Vice President of the - Standing - The minister - Worker

This is often used at work , For example, some tables store data about organizational relationships , Some of them are the menu on the left , If you need to use the organization table in a complex business, you will usually use this .

For the above Syntax , There's a keyword in it prior, This can be on the left and right of the equal sign ,level You can have it or not , It is only used to indicate which level the row data belongs to .

   select t.*,level from tablename t       
   start with  Conditions 1
   connect by   prior  Subfield id= Parent field id
   where  Conditions 3;

If it's on the left , The query is the leaf node based on condition one ;
If it's on the right , The upper nodes based on condition one are checked .

On the left

from 1001 Start , Query its child nodes ;

select t.*,level from bank t
start with t.funcid=1001
connect by prior t.funcid=t.parentid
order by funcid;

  On the right

check 1001 The parent node of the start , So there's only one piece of data ;

select t.*, level
  from bank t
 start with t.funcid = 1001
connect by t.funcid = prior t.parentid
 order by funcid;

6、 ... and .keep function

keep yes oracle An analysis function under ,Oracle In order to solve the problem of querying the maximum value in the subset , Put forward KEEP() grammar , That is to say Take the same group and sort by a certain field , Take the minimum or maximum value for the specified field .

The grammar is as follows ：

min | max(column1) keep (dense_rank first | last order by column2) over (partion by column3);

Examples are as follows ： Build a watch , Use the top with as Create a temporary table as follows ：

WITH workers AS(  
  SELECT 'DOM1' dept, 'zhangsan' names , 23 age, 4000 salaries FROM dual UNION ALL  
  SELECT 'DOM1' dept, 'lisi' names     , 35 age, 9000 salaries FROM dual UNION ALL  
  SELECT 'DOM2' dept, 'wangwu' names   , 26 age, 6500 salaries FROM dual UNION ALL  
  SELECT 'DOM2' dept, 'maliu' names    , 28 age, 6000 salaries FROM dual UNION ALL  
  SELECT 'DOM2' dept, 'zhaoqi' names   , 26 age, 5000 salaries FROM dual UNION ALL  
  SELECT 'DOM1' dept, 'liba' names     , 23 age, 3000 salaries FROM dual   
)
select * from workers;

Enquiries are as follows

​ 

Business objectives ： Get the youngest person in the Department , A record of highest wages ;

Then you can use keep(dense_rank first/last) To deal with it , Replace the above sf..works, such as 

SELECT w.dept,
       max(w.salaries) KEEP(dense_rank first order by w.age) max_salary
  FROM workers w
 GROUP BY dept;

​

intend Check the largest salary by age , Because this is the youngest , So use the first, If the oldest , use last;

explain

1.keep It means “ keep ”, Records that meet the conditions in parentheses will be maintained , use order by  There will naturally be first and last 了 , This is also a fixed way of writing ;

2.dense_rank It is a sort policy , This is generally the default ;

3.first/last Is to filter the data , Here we sift age Minimum record ;

4. Due to the use of aggregate functions , So finally we need to have group by;

7、 ... and .over function

grammar ：rank()/dense_rank over(partition by A order by B)

over Is an analytic function , According to the fields A Partition the results , Sort by field within each partition ; 
over Not to be used alone , Need and row_number(),rank() and dense_rank,lag() and lead(),sum() And so on

explain ：

  •            over() On what conditions ; 
  •            partition by Group by which field ;
  •            order by Sort by which field ;

Be careful ：

  •         Use rank()/dense_rank() when , You have to bring order by Otherwise it's illegal ;
  •         rank(): Jump order , If there are two first levels , Next is the third level .
  • 　　 dense_rank(): Sequential order , If there are two first levels , Next is still the second level .

for instance ：

Create a table and insert data ：

create table EMP  
(  
  empno    NUMBER(4) not null,  
  ename    VARCHAR2(10),  
  job      VARCHAR2(9),  
  mgr      NUMBER(4),  
  hiredate DATE,  
  sal      NUMBER(7,2),  
  comm     NUMBER(7,2),  
  deptno   NUMBER(2)  
)  
alter table EMP  
  add constraint PK_EMP primary key (EMPNO);  
  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7369, 'SMITH', 'CLERK', 7902, to_date('17-12-1980', 'dd-mm-yyyy'), 800, null, 20);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7499, 'ALLEN', 'SALESMAN', 7698, to_date('20-02-1981', 'dd-mm-yyyy'), 1600, 300, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7521, 'WARD', 'SALESMAN', 7698, to_date('22-02-1981', 'dd-mm-yyyy'), 1250, 500, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7566, 'JONES', 'MANAGER', 7839, to_date('02-04-1981', 'dd-mm-yyyy'), 2975, null, 20);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7654, 'MARTIN', 'SALESMAN', 7698, to_date('28-09-1981', 'dd-mm-yyyy'), 1250, 1400, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7698, 'BLAKE', 'MANAGER', 7839, to_date('01-05-1981', 'dd-mm-yyyy'), 2850, null, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7782, 'CLARK', 'MANAGER', 7839, to_date('09-06-1981', 'dd-mm-yyyy'), 2450, null, 10);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7788, 'SCOTT', 'ANALYST', 7566, to_date('19-04-1987', 'dd-mm-yyyy'), 3000, null, 20);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7839, 'KING', 'PRESIDENT', null, to_date('17-11-1981', 'dd-mm-yyyy'), 5000, null, 10);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7844, 'TURNER', 'SALESMAN', 7698, to_date('08-09-1981', 'dd-mm-yyyy'), 1500, 0, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7876, 'ADAMS', 'CLERK', 7788, to_date('23-05-1987', 'dd-mm-yyyy'), 1100, null, 20);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7900, 'JAMES', 'CLERK', 7698, to_date('03-12-1981', 'dd-mm-yyyy'), 950, null, 30);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7902, 'FORD', 'ANALYST', 7566, to_date('03-12-1981', 'dd-mm-yyyy'), 3000, null, 20);  
insert into EMP (empno, ename, job, mgr, hiredate, sal, comm, deptno)   
       values (7934, 'MILLER', 'CLERK', 7782, to_date('23-01-1982', 'dd-mm-yyyy'), 1300, null, 10);

The query is shown in the figure ：

 1、rank()/dense_rank() over(partition by ...order by ...)

Business one ： Ask for information about the highest paid employees in each department

The common expression is ：

select *
  from (select ename     full name ,
               job       occupation ,
               hiredate  Date of entry ,
               e.sal     Wages ,
               e.deptno  department 
          from emp e,
               (select deptno, max(sal) sal from emp group by deptno) t
         where e.deptno = t.deptno
           and e.sal = t.sal)
 order by  department ;

 

Use over Function is ：

select empno, ename, job, hiredate, sal, deptno
  from (select empno,
               ename,
               job,
               hiredate,
               sal,
               deptno,
               rank() over(partition by deptno order by sal desc) r
          from emp)
 where r = 1;

select empno, ename, job, hiredate, sal, deptno
  from (select empno,
               ename,
               job,
               hiredate,
               sal,
               deptno,
               dense_rank() over(partition by deptno order by sal desc) r
          from emp)
 where r = 1

  The results of these two operations are consistent with the above ,
By Department (deptno) After partition, according to the salary (sal) Sort from top to bottom , Then each district only takes one result, that is, the maximum wage ;

  another ： If you want to inquire about the minimum wage ,desc Change it to asc that will do ;

2、min()/max() over(partition by ...)

Business II ： Query employee information and calculate employee salary and department maximum / The difference in the minimum wage

Be careful ：min/max no need order by

The common expression is ：

select ename  full name ,
       job  occupation ,
       hiredate  Date of entry ,
       e.deptno  department ,
       e.sal  Wages ,
       e.sal - me.min_sal  Minimum difference ,
       me.max_sal - e.sal  Maximum difference 
  from emp e,
       (select deptno, min(sal) min_sal, max(sal) max_sal
          from emp
         group by deptno) me
 where e.deptno = me.deptno
 order by e.deptno, e.sal;

Use over by ：

select ename  full name ,
       job  occupation ,
       hiredate  Date of entry ,
       deptno  department ,
       sal  Wages ,
       min(sal) over(partition by deptno)  Departmental minimum wage ,
       max(sal) over(partition by deptno)  The highest salary in the Department ,
       nvl(sal - min(sal) over(partition by deptno), 0)  Departmental minimum wage difference ,
       nvl(max(sal) over(partition by deptno) - sal, 0)  The maximum wage difference of the Department 
  from emp
 order by deptno, sal;

 ​​​​​

Inquire about Difference The results are consistent with the common , And less than the code above ;

If you look at the code scanning results on both sides ？ As shown in the figure below ：

  And the bottom only went through a full table scan

So to sum up the comparison , It can be optimized according to business requirements .

Say something else ：

Why not order by sal 了 ？

order by The default is asc Sort from small to large , Write the query as follows 

select ename  full name ,
       job  occupation ,
       hiredate  Date of entry ,
       deptno  department ,
       sal  Wages ,
       min(sal) over(partition by deptno order by sal)  Departmental minimum wage ,
       max(sal) over(partition by deptno order by sal)  The highest salary in the Department ,
       nvl(sal - min(sal) over(partition by deptno order by sal), 0)  Departmental minimum wage difference ,
       nvl(max(sal) over(partition by deptno order by sal) - sal, 0)  The maximum wage difference of the Department 
  from emp
 order by deptno, sal;

 

The personal salary is actually the same as the maximum salary of the Department , Then the difference is 0, The reason is not clear , Remember to use min/max Wait for the aggregate function over The field of this aggregation cannot be used in order by, That means you don't have to write order by 了 .

 3、lead()/lag() over(partition by ... order by ...)

  First of all lead function , No leading;

These two functions , Is an offset function ;
Its purpose is ： You can find out the next or previous value corresponding to each line of records in the same field .

lead() It's after taking N Row data 

select ename  full name ,
       job  occupation ,
       hiredate  Date of entry ,
       deptno  department ,
       sal  Wages ,
       lead(sal) over(order by deptno)  Offset 1,
       lead(sal,2) over(order by deptno)  Offset 2
  from emp
 order by deptno, sal;

Pictured , For example, the offset is 1 It's from The first 2 OK, we're going to start with 14 That's ok this N Row data , So it is called after taking N That's ok ;

  • lead(column,num,flag)
  • col_name Is the column name ;num Is the value below the orientation ;flag It's a sign , That is, if the lower value is null, it will be taken as flag;
  • for example lead(column,1,null) This is to take a value down , If this value is empty, it will be calculated as null , Of course, you can also replace with other values .

lag() It's before taking N Row data 

select ename  full name ,
       job  occupation ,
       hiredate  Date of entry ,
       deptno  department ,
       sal  Wages ,
       lag(sal) over(order by deptno)  Offset 1,
       lag(sal,2,0) over(order by deptno)  Offset 2
  from emp
 order by deptno, sal;

Pictured , For example, the offset is 1 It's from The first 13 OK, we're going to start with 1 That's ok this N Row data , So it's called "before taking" N That's ok ;

Okay , Now let's explain the business problem ：

Business three ： Calculate the personal salary under the same department and one higher than yourself / The difference of one lower salary 

select ename  full name ,
       job  occupation ,
       sal  Wages ,
       deptno  department ,
       lead(sal, 1, 0) over(partition by deptno order by sal)  The last person whose salary is higher than his own ,
       lag(sal, 1, 0) over(partition by deptno order by sal)  The last person whose salary is lower than his own ,
       nvl(lead(sal) over(partition by deptno order by sal) - sal, 0)  A difference higher than one's salary ,
       nvl(sal - lag(sal) over(partition by deptno order by sal), 0)  A difference lower than one's salary 
  from emp;

  The function has been described , Do not explain ;

Then list some common ways to write collocation aggregate functions 

select ename  full name , job  occupation , sal  Wages , deptno  department ,   
    first_value(sal) over(partition by deptno) first_sal,   
    last_value(sal) over(partition by deptno) last_sal,  
    min(sal) over(partition by deptno) min_sal,   
    max(sal) over(partition by deptno) max_sal,  
    sum(sal) over(partition by deptno)  Total Department salary ,   
    avg(sal) over(partition by deptno)  Department average wage ,   
    count(1) over(partition by deptno)  Total number of departments ,   
    row_number() over(partition by deptno order by sal)  Serial number    
 from emp;

Excerpt from Blog ​​​​​​​ And made a more detailed description

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