1 SQL cont.. 2 Outline Unions, intersections, differences (6.2.5, 6.4.2) Subqueries (6.3) Aggregations (6.4.3 – 6.4.6) Hint for reading the textbook:

Slides:



Advertisements
Similar presentations
SQL Introduction Standard language for querying and manipulating data Structured Query Language Many standards out there: SQL92, SQL2, SQL3. Vendors support.
Advertisements

Relational Algebra (end) SQL April 19 th, Complex Queries Product ( pid, name, price, category, maker-cid) Purchase (buyer-ssn, seller-ssn, store,
1 Lecture 12: SQL Friday, October 26, Outline Simple Queries in SQL (5.1) Queries with more than one relation (5.2) Subqueries (5.3) Duplicates.
1 Lecture 03: Advanced SQL. 2 Outline Unions, intersections, differences Subqueries, Aggregations, NULLs Modifying databases, Indexes, Views Reading:
M.P. Johnson, DBMS, Stern/NYU, Spring C : Database Management Systems Lecture #11 M.P. Johnson Stern School of Business, NYU Spring, 2008.
SQL April 25 th, Agenda Grouping and aggregation Sub-queries Updating the database Views More on views.
1 Lecture 03: SQL Friday, January 7, Administrivia Have you logged in IISQLSRV yet ? HAVE YOU CHANGED YOUR PASSWORD ? Homework 1 is now posted.
M.P. Johnson, DBMS, Stern/NYU, Spring C : Database Management Systems Lecture #12 M.P. Johnson Stern School of Business, NYU Spring, 2005.
1 Lecture 02: Basic SQL. 2 Outline Data in SQL Simple Queries in SQL Queries with more than one relation Reading: Chapter 3, “Simple Queries” from SQL.
3) Student (ufid, major) Quiz(Q_num, point_pos) Q_score(q_num,ufid,points_scored) (in TRC) A) Which quizzes did a “CE” major score 100% on? { a[Q_num]:
M.P. Johnson, DBMS, Stern/NYU, Sp20041 C : Database Management Systems Lecture #12 Matthew P. Johnson Stern School of Business, NYU Spring, 2004.
Correlated Queries SELECT title FROM Movie AS Old WHERE year < ANY (SELECT year FROM Movie WHERE title = Old.title); Movie (title, year, director, length)
1 Lecture 3: More SQL Friday, January 9, Agenda Homework #1 on the web site today. Sign up for the mailing list! Next Friday: –In class ‘activity’
Union, Intersection, Difference (SELECT name FROM Person WHERE City=“Seattle”) UNION (SELECT name FROM Person, Purchase WHERE buyer=name AND store=“The.
1 Information Systems Chapter 6 Database Queries.
Complex Queries (1) Product ( pname, price, category, maker)
CSE544: SQL Monday 3/27 and Wednesday 3/29, 2006.
Integrity Constraints An important functionality of a DBMS is to enable the specification of integrity constraints and to enforce them. Knowledge of integrity.
+ From Relational Algebra to SQL W2013 CSCI 2141.
Lecture 3: Introduction to SQL September 29, 2014.
Exercises Product ( pname, price, category, maker) Purchase (buyer, seller, store, product) Company (cname, stock price, country) Person( per-name, phone.
1 Lecture 4: More SQL Monday, January 13th, 2003.
SQL (almost end) April 26 th, Agenda HAVING clause Views Modifying views Reusing views.
SQL April 22 th, Agenda Union, intersections Sub-queries Modifying the database Views Modifying views Reusing views.
IM433-Industrial Data Systems Management Lecture 5: SQL.
More SQL: Complex Queries, Triggers, Views, and Schema Modification UMM AL QURA UNIVERSITY College of Computer Dr. Ali Al Najjar 1.
Lectures 2&3: Introduction to SQL. Lecture 2: SQL Part I Lecture 2.
SQL SQL Review. SQL Introduction Standard language for querying and manipulating data Structured Query Language Many standards out there: ANSI SQL, SQL92.
SQL. SQL Introduction Standard language for querying and manipulating data Structured Query Language Many standards out there: ANSI SQL, SQL92 (a.k.a.
1 SQL Additional Notes. 2  1 Group and Aggregation*  2 Execution Order*  3 Join*  4 Find the maximum  5 Line Format SQL Additional Notes *partially.
1 Introduction to Database Systems CSE 444 Lecture 02: SQL September 28, 2007.
1 Introduction to Database Systems CSEP544 Lecture #1 January 5, 2007.
1 Lecture 02: SQL Friday, September 30, Administrivia Homework 1 is out. Due: Wed., Oct. 12 Did you login on IISQLSRV ? Did you change your password.
1 Introduction to Database Systems CSEP 544 Lecture #1 March 29, 2004 Alon Halevy.
Aggregation SELECT Sum(price) FROM Product WHERE manufacturer=“Toyota” SQL supports several aggregation operations: SUM, MIN, MAX, AVG, COUNT Except COUNT,
1 Lecture 03: SQL Monday, January 9, Project t/Default.aspxhttp://iisqlsrv.cs.washington.edu/444/Projec.
SQL. SQL Introduction Standard language for querying and manipulating data Structured Query Language Many standards out there: ANSI SQL, SQL92 (a.k.a.
SQL.
Cours 7: Advanced SQL.
Lecture 04: SQL Monday, January 10, 2005.
Database Systems Subqueries, Aggregation
Cse 344 April 4th – Subqueries.
Monday 3/27 and Wednesday 3/29, 2006
Class 2 Relational Data Languages
Lecture 2 (cont’d) & Lecture 3: Advanced SQL – Part I
Cse 344 January 12th –joins.
Introduction to Database Systems CSE 444 Lecture 03: SQL
January 17th – Subqueries
Lecture 4: Advanced SQL – Part II
Introduction to SQL Wenhao Zhang October 5, 2018.
Introduction to Database Systems CSE 444 Lecture 03: SQL
CSE544 SQL Wednesday, March 31, 2004.
Building a Database Application
SQL Introduction Standard language for querying and manipulating data
Lecture 12: SQL Friday, October 20, 2000.
Lectures 3: Introduction to SQL Part II
Introduction to Database Systems CSE 444 Lecture 02: SQL
Lectures 5: Introduction to SQL 4
Lecture 4: SQL Thursday, January 11, 2001.
Lectures 6: Introduction to SQL 5
Integrity Constraints
Lecture 3 Monday, April 8, 2002.
Class 2 Relational Data Languages
Lecture 4: SQL Wednesday, April 10, 2002.
Lecture 03: SQL Friday, October 3, 2003.
Lecture 3: Relational Algebra and SQL
Lecture 04: SQL Monday, October 6, 2003.
Lecture 14: SQL Wednesday, October 31, 2001.
Presentation transcript:

1 SQL cont.

2 Outline Unions, intersections, differences (6.2.5, 6.4.2) Subqueries (6.3) Aggregations (6.4.3 – 6.4.6) Hint for reading the textbook: read the entire chapter 6 ! Recommended reading from “SQL for Nerds”: chapter 4, “More complex queries” (you will find it very useful for subqueries)

3 Cartesian Product GameColor Baketball1 Footbal2 Tennis1 SELECT * FROM Games SELECT * FROM Games Color_IDColor_Name 1RED 2GREEN SELECT * FROM Colors SELECT * FROM Colors

4 Cartesian Product GameColorColor_IDColor_Name Baketball11RED Baketball12GREEN Footbal21RED Footbal22GREEN Tennis11RED Tennis12GREEN SELECT * FROM Games, Colors SELECT * FROM Games, Colors

5 Cartesian Product GameColorColor_IDColor_Name Baketball11RED Baketball12GREEN Footbal21RED Footbal22GREEN Tennis11RED Tennis12GREEN SELECT * FROM Games, Colors WHERE Color = Color_ID SELECT * FROM Games, Colors WHERE Color = Color_ID

6 Cartesian Product GameColorColor_IDColor_Name Baketball11RED Footbal22GREEN Tennis11RED SELECT * FROM Games, Colors WHERE Color = Color_ID SELECT * FROM Games, Colors WHERE Color = Color_ID

7 Renaming Columns PNamePriceCategoryManufacturer Gizmo$19.99GadgetsGizmoWorks Powergizmo$29.99GadgetsGizmoWorks SingleTouch$149.99PhotographyCanon MultiTouch$203.99HouseholdHitachi SELECT Pname AS prodName, Price AS askPrice FROM Product WHERE Price > 100 Product prodNameaskPrice SingleTouch$ MultiTouch$ Query with renaming

8 Union, Intersection, Difference (SELECT name FROM Person WHERE City=“Seattle”) UNION (SELECT name FROM Person, Purchase WHERE buyer=name AND store=“The Bon”) (SELECT name FROM Person WHERE City=“Seattle”) UNION (SELECT name FROM Person, Purchase WHERE buyer=name AND store=“The Bon”) Similarly, you can use INTERSECT and EXCEPT. You must have the same attribute names (otherwise: rename)

9 (SELECT DISTINCT R.A FROM R) INTERSECT ( (SELECT S.A FROM S) UNION (SELECT T.A FROM T)) (SELECT DISTINCT R.A FROM R) INTERSECT ( (SELECT S.A FROM S) UNION (SELECT T.A FROM T))

10 Conserving Duplicates (SELECT name FROM Person WHERE City=“Seattle”) UNION ALL (SELECT name FROM Person, Purchase WHERE buyer=name AND store=“The Bon”) (SELECT name FROM Person WHERE City=“Seattle”) UNION ALL (SELECT name FROM Person, Purchase WHERE buyer=name AND store=“The Bon”)

11 Subqueries (Static..) A subquery producing a single value: In this case, the subquery returns one value. If it returns more, it’s a run-time error SELECT Purchase.product FROM Purchase WHERE buyer = (SELECT name FROM Person WHERE ssn = ‘ ‘); SELECT Purchase.product FROM Purchase WHERE buyer = (SELECT name FROM Person WHERE ssn = ‘ ‘);

12 Subqueries Returning Relations SELECT Company.name FROM Company, Product WHERE Company.name=Product.maker AND Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyer = ‘Joe Blow‘); SELECT Company.name FROM Company, Product WHERE Company.name=Product.maker AND Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyer = ‘Joe Blow‘); Find companies who manufacture products bought by Joe Blow. Here the subquery returns a set of values: no more runtime errors

13 Subqueries Returning Relations SELECT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ Equivalent to: Is this query equivalent to the previous one ? Beware of duplicates !

14 Removing Duplicates SELECT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT DISTINCT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT DISTINCT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’  Multiple copies  Single copies

15 Removing Duplicates SELECT DISTINCT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT DISTINCT Company.name FROM Company, Product, Purchase WHERE Company.name= Product.maker AND Product.name = Purchase.product AND Purchase.buyer = ‘Joe Blow’ SELECT DISTINCT Company.name FROM Company, Product WHERE Company.name= Product.maker AND Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyer = ‘Joe Blow’) SELECT DISTINCT Company.name FROM Company, Product WHERE Company.name= Product.maker AND Product.name IN (SELECT Purchase.product FROM Purchase WHERE Purchase.buyer = ‘Joe Blow’) Now they are equivalent

16 Subqueries Returning Relations SELECT name FROM Product WHERE price > ALL (SELECT price FROM Product WHERE maker=‘Gizmo-Works’) SELECT name FROM Product WHERE price > ALL (SELECT price FROM Product WHERE maker=‘Gizmo-Works’) Product ( pname, price, category, maker) Find products that are more expensive than all those produced By “Gizmo-Works” You can also use: s > ALL R s > ANY R EXISTS R

17 Question for Database Fans and their Friends Can we express this query as a single SELECT- FROM-WHERE query, without subqueries ? Hint: show that all SFW queries are monotone (figure out what this means). A query with ALL is not monotone

18 Conditions on Tuples SELECT DISTINCT Company.name FROM Company, Product WHERE Company.name= Product.maker AND (Product.name,price) IN (SELECT Purchase.product, Purchase.price) FROM Purchase WHERE Purchase.buyer = “Joe Blow”); SELECT DISTINCT Company.name FROM Company, Product WHERE Company.name= Product.maker AND (Product.name,price) IN (SELECT Purchase.product, Purchase.price) FROM Purchase WHERE Purchase.buyer = “Joe Blow”); May not work in Mysql...

19 Correlated Queries SELECT DISTINCT title FROM Movie AS x WHERE year <> ANY (SELECT year FROM Movie WHERE title = x.title); SELECT DISTINCT title FROM Movie AS x WHERE year <> ANY (SELECT year FROM Movie WHERE title = x.title); Movie (title, year, director, length) Find movies whose title appears more than once. Note (1) scope of variables (2) this can still be expressed as single SFW correlation

20 Complex Correlated Query Product ( pname, price, category, maker, year) Find products (and their manufacturers) that are more expensive than all products made by the same manufacturer before 1972 Powerful, but much harder to optimize ! SELECT DISTINCT pname, maker FROM Product AS x WHERE price > ALL (SELECT price FROM Product AS y WHERE x.maker = y.maker AND y.year < 1972); SELECT DISTINCT pname, maker FROM Product AS x WHERE price > ALL (SELECT price FROM Product AS y WHERE x.maker = y.maker AND y.year < 1972);

21 Aggregation SELECT Avg(price) FROM Product WHERE maker=“Toyota” SELECT Avg(price) FROM Product WHERE maker=“Toyota” SQL supports several aggregation operations: SUM, MIN, MAX, AVG, COUNT

22 Aggregation: Count SELECT Count(*) FROM Product WHERE year > 1995 SELECT Count(*) FROM Product WHERE year > 1995 Except COUNT, all aggregations apply to a single attribute

23 Aggregation: Count COUNT applies to duplicates, unless otherwise stated: SELECT Count(category) same as Count(*) FROM Product WHERE year > 1995 Better: SELECT Count(DISTINCT category) FROM Product WHERE year > 1995

24 Simple Aggregation Purchase(product, date, price, quantity) Example 1: find total sales for the entire database SELECT Sum(price * quantity) FROM Purchase Example 1’: find total sales of bagels SELECT Sum(price * quantity) FROM Purchase WHERE product = ‘bagel’

25 Simple Aggregations Purchase

26 Grouping and Aggregation Usually, we want aggregations on certain parts of the relation. Purchase(product, date, price, quantity) Example 2: find total sales after 10/1 per product. SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUPBY product SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUPBY product Let’s see what this means…

27 Grouping and Aggregation 1. Compute the FROM and WHERE clauses. 2. Group by the attributes in the GROUPBY 3. Select one tuple for every group (and apply aggregation) SELECT can have (1) grouped attributes or (2) aggregates.

28 First compute the FROM-WHERE clauses (date > “10/1”) then GROUP BY product:

29 Then, aggregate SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUPBY product SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUPBY product

30 GROUP BY v.s. Nested Queries SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUP BY product SELECT product, Sum(price*quantity) AS TotalSales FROM Purchase WHERE date > “10/1” GROUP BY product SELECT DISTINCT x.product, (SELECT Sum(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.date > ‘10/1’) AS TotalSales FROM Purchase x WHERE x.date > “10/1” SELECT DISTINCT x.product, (SELECT Sum(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.date > ‘10/1’) AS TotalSales FROM Purchase x WHERE x.date > “10/1”

31 Another Example SELECT product, Sum(price * quantity) AS SumSales Max(quantity) AS MaxQuantity FROM Purchase GROUP BY product SELECT product, Sum(price * quantity) AS SumSales Max(quantity) AS MaxQuantity FROM Purchase GROUP BY product For every product, what is the total sales and max quantity sold?

32 HAVING Clause SELECT product, Sum(price * quantity) FROM Purchase WHERE date > “9/1” GROUP BY product HAVING Sum(quantity) > 100 SELECT product, Sum(price * quantity) FROM Purchase WHERE date > “9/1” GROUP BY product HAVING Sum(quantity) > 100 Same query, except that we consider only products that had at least 100 buyers. HAVING clause contains conditions on aggregates.

33 General form of Grouping and Aggregation SELECT S FROM R 1,…,R n WHERE C1 GROUP BY a 1,…,a k HAVING C2 S = may contain attributes a 1,…,a k and/or any aggregates but NO OTHER ATTRIBUTES C1 = is any condition on the attributes in R 1,…,R n C2 = is any condition on aggregate expressions Why ?

34 General form of Grouping and Aggregation SELECT S FROM R 1,…,R n WHERE C1 GROUP BY a 1,…,a k HAVING C2 Evaluation steps: 1.Compute the FROM-WHERE part, obtain a table with all attributes in R 1,…,R n 2.Group by the attributes a 1,…,a k 3.Compute the aggregates in C2 and keep only groups satisfying C2 4.Compute aggregates in S and return the result

35 Aggregation Author(login,name) Document(url, title) Wrote(login,url) Mentions(url,word)

36 Find all authors who wrote at least 10 documents: Attempt 1: with nested queries SELECT DISTINCT Author.name FROM Author WHERE count(SELECT Wrote.url FROM Wrote WHERE Author.login=Wrote.login) > 10 SELECT DISTINCT Author.name FROM Author WHERE count(SELECT Wrote.url FROM Wrote WHERE Author.login=Wrote.login) > 10 This is SQL by a novice

37 Find all authors who wrote at least 10 documents: Attempt 2: SQL style (with GROUP BY) SELECT Author.name FROM Author, Wrote WHERE Author.login=Wrote.login GROUP BY Author.name HAVING count(wrote.url) > 10 SELECT Author.name FROM Author, Wrote WHERE Author.login=Wrote.login GROUP BY Author.name HAVING count(wrote.url) > 10 This is SQL by an expert No need for DISTINCT: automatically from GROUP BY

38 Find all authors who have a vocabulary over words: SELECT Author.name FROM Author, Wrote, Mentions WHERE Author.login=Wrote.login AND Wrote.url=Mentions.url GROUP BY Author.name HAVING count(distinct Mentions.word) > SELECT Author.name FROM Author, Wrote, Mentions WHERE Author.login=Wrote.login AND Wrote.url=Mentions.url GROUP BY Author.name HAVING count(distinct Mentions.word) > Look carefully at the last two queries: you may be tempted to write them as a nested queries, but in SQL we write them best with GROUP BY