SQL Advanced in MS SQL: Difference between revisions

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* Grouping sets
* Grouping sets
* Hierarchical queries
* Hierarchical queries
* Analytic functions
* Insert
* Insert
* Delete
* Delete
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       (deptno, job, grade),
       (deptno, job, grade),
       (deptno, grade),  
       (deptno, grade),  
       (job, grade) )  
      (deptno, job),
      (job, grade),
       (job),
      (grade),
      (deptno)
      )  
   order by deptno, job, grade
   order by deptno, job, grade
;
;
Line 1,101: Line 1,107:
== Hierarchy ==
== Hierarchy ==
* Using '''outer joins''' only
* Using '''outer joins''' only
* Recursively with '''CTE''' (Common Table Expression)
* Recursively with '''CTE''' (''Common Table Expression'')
* Built-in support - '''hierarchyid''' column type  
* Built-in support - '''hierarchyid''' column type  


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</syntaxhighlight>
</syntaxhighlight>


=== CTE Exercise ===
=== CTE Ex Skeleton ===
<syntaxhighlight lang="sql">
<syntaxhighlight lang="sql">
-- Rewrite above 'subqueried' solution into 'recursive CTE'
-- Rewrite above 'join' solution into 'recursive CTE'
WITH ...
WITH ...
AS (SELECT ...
AS (SELECT ...
Line 1,135: Line 1,141:
;
;
</syntaxhighlight>
</syntaxhighlight>
<!-- Solution:
 
=== CTE Exercises ===
# Get the '''level for each node'''
# Get the '''path for each node'''
# Get the '''sorted hierarchy tree result'''
#* parent followed by all of its children, then all the children of each one of those and so on
#* use functions '''row_number()''' and '''power()'''
<!-- Solution Ex 1
WITH hierarchicalList
WITH hierarchicalList
AS (SELECT PoshBoss.id, PoshBoss.ename, PoshBoss.mgr, 0 AS EmpLev
AS (SELECT PoshBoss.id, PoshBoss.ename, PoshBoss.mgr, 0 AS EmpLev
Line 1,148: Line 1,161:
FROM hierarchicalList
FROM hierarchicalList
;
;
-->
<!-- Solution Ex2
with chL as (
    select id, ename, cast(ename as varchar(200)) as hierPath
    from emp where id = 7839
    union all
    select r.id, r.ename, cast(chL.hierPath + ' / ' + r.ename as varchar(200)) as hierPath
    from emp r inner join chL on chL.id = r.mgr
)
select * from chL;
-->
<!-- Solution Ex3
with chL as (
    select id, ename, cast(ename as varchar(200)) as hierPath, 0 as lvl,
    row_number() over (partition by mgr order by ename) / power(10.0,1) as sortNo
    from emp where id = 7839
    union all
    select r.id, r.ename, cast(chL.hierPath + ' / ' + r.ename as varchar(200)) as hierPath, chL.lvl+1 as lvl,
    chL.sortNo + row_number() over (partition by r.mgr order by r.ename) / power(10.0, chL.lvl+1) as sortNo
    from emp r inner join chL on chL.id = r.mgr
)
select * from chL
order by sortNo;
-->
-->


=== hierarchyid Example ===
=== hierarchyid ''Example & Exercises'' ===
[[SQL_Exercises#hierarchyid_%28mssql%29|hierarchyid]]
<!--
<syntaxhighlight lang="sql">
<syntaxhighlight lang="sql">
-- TODO
-- TODO
Line 1,159: Line 1,197:
-- TODO
-- TODO
</syntaxhighlight>
</syntaxhighlight>
-->
== Analytic functions ==
* <small>https://learn.microsoft.com/en-us/sql/t-sql/functions/analytic-functions-transact-sql?view=sql-server-ver16</small>
<syntaxhighlight lang="sql">
-- Example 1
---- Find the 2 highest paid employees in each department
SELECT deptno, ename, sal
FROM
(
  SELECT
    deptno, ename, sal,
    ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY sal desc) rn
  FROM emp
) AS context
WHERE rn <= 2
ORDER BY deptno, sal DESC;
-- Exercise 1
---- Count cumulative distribution of salaries by manager
------ only for people from departments 10 and 30
------ sort it by hierarchy and id
-- Exercise 2
---- Extend above with also percent rank
------ show department name
------ sort it within window by hierarchy
</syntaxhighlight>
<!-- Solutions
SELECT mgr, id, sal,
  CUME_DIST() OVER (PARTITION BY mgr ORDER BY sal) "Cum dist by Mgr"
  FROM emp
  WHERE deptno IN (10, 30)
  ORDER BY mgr, id;
SELECT e.deptno, ename, mgr, dname,
      CUME_DIST () OVER (PARTITION BY e.deptno ORDER BY mgr) AS CumeDist, 
      PERCENT_RANK() OVER (PARTITION BY e.deptno ORDER BY mgr ) AS PctRank 
FROM emp AS e
    INNER JOIN dept AS d   
    ON d.deptno = e.deptno
WHERE e.deptno IN (10, 30) 
ORDER BY e.deptno, mgr DESC;
-->
=== LAG, LEAD ===
<syntaxhighlight lang="sql">
------ LAG vs LEAD example - no need to use 'self-join'
--- For each Clerk find the salary of the employee hired just before
SELECT hiredate, ename, sal,
  LAG(sal, 1, 0) OVER (ORDER BY hiredate) AS prev_sal
  FROM emp
  WHERE job = 'CLERK'
  ORDER BY hiredate;
----------------------
-- Exercise
--- For each employee in Department 30 find the hire date of the employee hired just after
</syntaxhighlight>
<!-- Solution
SELECT hiredate, ename,
  LEAD(hiredate, 1) OVER (ORDER BY hiredate) AS "NextHired"
  FROM emp
  WHERE deptno = 30
  ORDER BY hiredate;
-->
=== FIRST, LAST ===
<syntaxhighlight lang="sql">
-- Return the employee with the fewest monies compared to other employees with the same job title
SELECT job,
    ename,
    sal,
    dname,
    FIRST_VALUE(ename) OVER (
        PARTITION BY job ORDER BY sal ASC ROWS UNBOUNDED PRECEDING
    ) AS FewestMonies
FROM emp AS e
INNER JOIN dept AS d
    ON e.deptno = d.deptno
ORDER BY job;
-------------------------
-- Exercise
---- Return the hire date of the last employee in each department for the given salary
------ show department name and monies grade
</syntaxhighlight>
<!-- Solution
SELECT e.deptno,
    ename,
    sal,
dname,
grade,
    hiredate,
    LAST_VALUE(hiredate) OVER (
        PARTITION BY e.deptno ORDER BY sal
    ) AS LastValue
FROM emp AS e
INNER JOIN dept AS d
    ON e.deptno = d.deptno
INNER JOIN salgrade AS s
    ON sal BETWEEN losal AND hisal
WHERE e.deptno IN (10, 30);
-->


== Example 1, 'WITH' ==
== Example 1, 'WITH' ==

Latest revision as of 15:14, 22 August 2024


SQL Advanced in MS SQL Training Materials

IMPORTANT NOTE for trainers only: 
Please DO NOT change these materials, especially the EXERCISES, without direct permission from LUKASZ SOKOLOWSKI.

Recap of Structured Query Language

  • DDL
  • DML

DDL

Data Definition Language

  • A collection of imperative verbs
  • Modify the schema of the database
  • Adding, changing, or deleting definitions of tables or other objects

DDL Statements

Examples

  • CREATE ...
    • FUNCTION, DATABASE, PROCEDURE, TYPE, USER, VIEW, SCHEMA
  • ALTER ...
    • INDEX, TABLE, TRIGGER
  • DROP ...
    • RULE, STATISTICS, SYNONYM
  • Other DDL statements
    • ENABLE TRIGGER - DISABLE TRIGGER - TRUNCATE TABLE - UPDATE STATISTICS
  • More here - learn.microsoft.com/en-us/sql/t-sql/statements/statements

DML

Data Manipulation Language

  • Used to insert, query, delete and update data in a database
  • Comprises the 'SQL-data change' statements
    • Modify stored data but not the schema or database objects

DML Statements

  • SELECT
  • INSERT
  • DELETE
  • UPDATE
  • MERGE

Exercises

  • Escape in LIKE, more special characters
  • CASE expression
  • Grouping sets
  • Hierarchical queries
  • Analytic functions
  • Insert
  • Delete
  • Update
  • Stored procedures
  • Transactions
  • Cursors
  • Triggers
  • Procedural programming, handling errors

Escape in LIKE, more special characters

--- DB preps
USE mike
GO

CREATE TABLE [dbo].[dept2](
	[DEPTNO] [numeric](2, 0) NOT NULL,
	[DNAME] [char](14) NULL,
	[LOC] [char](13) NULL,
)

INSERT INTO dept2(DEPTNO, DNAME, LOC)
VALUES (10, 'ACCOUNTING', 'NEW_YORK');

INSERT INTO dept2(DEPTNO, DNAME, LOC)
VALUES (20, 'RESEARCH', 'OLD_TOWN');

INSERT INTO dept2(DEPTNO, DNAME, LOC)
VALUES (30, 'SALES', 'PARIS');

INSERT INTO dept2(DEPTNO, DNAME, LOC)
VALUES (40, 'OPERATIONS', 'BERLIN');

--- Escape string in LIKE clause
SELECT
    dname, loc
FROM
    dept2
WHERE
    loc LIKE '%\_YORK' ESCAPE '\'
;

More specials in LIKE

  • [e-k] range
  • [bs] set
  • [^] negation (not) of range or set
  • [_], [%] simpler escaping
  • Examples
SELECT dname, loc FROM dept2
WHERE
    loc LIKE '%[_]%'
    -- dname LIKE '[a-o]%'
    -- dname LIKE '[an]%'
    -- dname LIKE '[^or]%'
;
  • Exercises
    • Find all employee names containing letters L or N or W
    • Find all employee names containing letters L and N and W
    • Search for people hired not in months: Jan and Dec

CASE expression

  • Example
 
---- Case examples
-- Simple vs Searched 
SELECT DISTINCT job, sal,
  ---- Simple - same condition
  (CASE grade 
    WHEN 5 THEN 'Bossy'
    WHEN 4 THEN 'CEO monies'
    ELSE 'Peanuts' END) AS How_Much,
  ---- Searched - multiple conditions
  (CASE
    WHEN sal BETWEEN 500 AND 1200 THEN 'Rice'
    WHEN ((sal BETWEEN 1250 AND 1800) AND comm IS NULL ) THEN 'Beans'
    ELSE 'Coconuts' END) AS "Who's rich?"
  FROM emp JOIN salgrade
    ON sal BETWEEN losal AND hisal
  ORDER BY 2;
---------------------------------
---- https://docs.oracle.com/cd/B19306_01/server.102/b14200/expressions004.htm
  • Exercise - Calculate report per job, like below
    • summarize salary bigger than 2000, otherwise put 0
    • average of commissions lower than 1000, otherwise 0
    • how many people
-- 5 rows
----------------------------------------------------------------------
Job        Summarized Salary    Average Commission    Number of people
----------------------------------------------------------------------
ANALYST	   6000                 0                     2
CLERK	   0                    0                     4
SALESMAN   0	                200                   4
MANAGER	   8275                 0                     3
PRESIDENT  5000                 0                     1
----------------------------------------------------------------------

Grouping sets

------ EXAMPLE - GROUPING SETS
---- Exercises
-- 1. Compare it with simple grouping (undo the comment)
-- 2. Improve it - group also by 'grade' and count it
----
select deptno, job, sum(sal)
  from emp
  where deptno in (20, 30)
  -- group by deptno, job
  group by GROUPING SETS( 
      (deptno, job), 
      (deptno), 
      (job) ) 
  order by deptno, job
;

Hierarchy

  • Using outer joins only
  • Recursively with CTE (Common Table Expression)
  • Built-in support - hierarchyid column type

Outer joins example

-- Show all workers and their bosses, within 3 levels of hierarchy.
SELECT W.ename AS Level0, B.ename AS Level1, B2.ename AS Level2, B3.ename AS Level3
FROM emp W LEFT JOIN emp B	ON W.mgr = B.id
		   LEFT JOIN emp B2 ON B.mgr = B2.id
		   LEFT JOIN emp B3 ON B2.mgr = B3.id
ORDER BY 4, 3, 2
;

CTE Example

-- Steps
---- 1. CTE, used in recursion
---- 2. UNION ALL
---- 3. INNER JOIN

CTE Ex Skeleton

-- Rewrite above 'join' solution into 'recursive CTE'
WITH ...
AS (SELECT ...
	UNION ALL
	SELECT ...
	)
SELECT *
FROM ...
;

CTE Exercises

  1. Get the level for each node
  2. Get the path for each node
  3. Get the sorted hierarchy tree result
    • parent followed by all of its children, then all the children of each one of those and so on
    • use functions row_number() and power()

hierarchyid Example & Exercises

hierarchyid

Analytic functions

-- Example 1
---- Find the 2 highest paid employees in each department
SELECT deptno, ename, sal
FROM
(
  SELECT
    deptno, ename, sal,
    ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY sal desc) rn
  FROM emp
) AS context
WHERE rn <= 2
ORDER BY deptno, sal DESC;

-- Exercise 1
---- Count cumulative distribution of salaries by manager
------ only for people from departments 10 and 30
------ sort it by hierarchy and id

-- Exercise 2
---- Extend above with also percent rank
------ show department name
------ sort it within window by hierarchy

LAG, LEAD

------ LAG vs LEAD example - no need to use 'self-join'
--- For each Clerk find the salary of the employee hired just before
SELECT hiredate, ename, sal,
  LAG(sal, 1, 0) OVER (ORDER BY hiredate) AS prev_sal
  FROM emp
  WHERE job = 'CLERK'
  ORDER BY hiredate;
----------------------
-- Exercise
--- For each employee in Department 30 find the hire date of the employee hired just after

FIRST, LAST

-- Return the employee with the fewest monies compared to other employees with the same job title
SELECT job,
    ename,
    sal,
    dname,
    FIRST_VALUE(ename) OVER (
        PARTITION BY job ORDER BY sal ASC ROWS UNBOUNDED PRECEDING
    ) AS FewestMonies
FROM emp AS e
INNER JOIN dept AS d
    ON e.deptno = d.deptno
ORDER BY job;
-------------------------
-- Exercise
---- Return the hire date of the last employee in each department for the given salary
------ show department name and monies grade

Example 1, 'WITH'

WITH functions_names as (
    SELECT 'GetAllTheBookingForms' f union all
    SELECT 'PrepareDelegatesDetails' f union all
    SELECT 'HelperFunction' f
)
SELECT
    f "Function name",
    lower( f )
    as "FN after"
FROM functions_names;

Exercise 1, 'INSERT' and 'WITH'

  • Create a table (FinancialReport): Ename, Dname, Grade
  • Insert into it all the related data - only from Sales and Accounting
  • Do it via WITH

Exercise 2, 'DELETE' and 'OUTPUT'

  • Remove grades 3 and 1 from FinancialReport
  • Save the deleted rows in a variable (temporary table)

Exercise 3, 'UPDATE'

  • Actualize grade 4: boost it up to 4k, make sure the rest of the data is consistent
  • Compare the old and new values in temporary variable (table)

Example 4, 'Stored Procedures', with 'INPUT' and 'OUTPUT' accordingly

Exercise 4a, 'Stored Procedures' with 'INPUT'

  • Prepare the procedure which will search for 'job' and 'sal'

Exercise 4b, 'Stored Procedures' with 'OUTPUT'

  • Invent the procedure which will return name and all the possible monies the person can earn in next 5 years.

Example 5, 'Transactions'

Exercise 5, 'Transactions'

  • Create simple named transaction which rises all the commissioned guys with 100 quid.

Example 6, 'Cursors'

Exercise 6, 'Fetch'

  • Fetch people with names ending with N or R.

Example 7, 'Triggers'

Exercise 7, 'Trigger' and 'DDL'

  • Run the above example and test it (=

Example 8, 'Procedural Programming'

Exercise 8, 'Procedural Programming'

  • Delete person only if he/she has no subordinates (use procedure and transaction with the save point)