1 SQL cont.. 2 Outline Unions, intersections, differences (6.2.5, 6.4.2) Subqueries (6.3)...
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Transcript of 1 SQL cont.. 2 Outline Unions, intersections, differences (6.2.5, 6.4.2) Subqueries (6.3)...
![Page 1: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/1.jpg)
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SQL cont.
![Page 2: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/2.jpg)
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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, “Morecomplex queries” (you will find it very useful for subqueries)
![Page 3: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/3.jpg)
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Cartesian Product
Game Color
Baketball 1
Footbal 2
Tennis 1
SELECT *FROM Games
SELECT *FROM Games
Color_IDColor_Nam
e
1 RED
2 GREEN
SELECT *FROM Colors
SELECT *FROM Colors
![Page 4: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/4.jpg)
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Cartesian Product
Game Color Color_ID Color_Name
Baketball 1 1 RED
Baketball 1 2 GREEN
Footbal 2 1 RED
Footbal 2 2 GREEN
Tennis 1 1 RED
Tennis 1 2 GREEN
SELECT *FROM Games, Colors
SELECT *FROM Games, Colors
![Page 5: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/5.jpg)
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Cartesian Product
Game Color Color_ID Color_Name
Baketball 1 1 RED
Baketball 1 2 GREEN
Footbal 2 1 RED
Footbal 2 2 GREEN
Tennis 1 1 RED
Tennis 1 2 GREEN
SELECT *FROM Games, ColorsWHERE Color = Color_ID
SELECT *FROM Games, ColorsWHERE Color = Color_ID
![Page 6: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/6.jpg)
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Cartesian Product
Game Color Color_ID Color_Name
Baketball 1 1 RED
Footbal 2 2 GREEN
Tennis 1 1 RED
SELECT *FROM Games, ColorsWHERE Color = Color_ID
SELECT *FROM Games, ColorsWHERE Color = Color_ID
![Page 7: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/7.jpg)
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Renaming Columns
PName Price Category Manufacturer
Gizmo $19.99 Gadgets GizmoWorks
Powergizmo $29.99 Gadgets GizmoWorks
SingleTouch $149.99 Photography Canon
MultiTouch $203.99 Household Hitachi
SELECT Pname AS prodName, Price AS askPriceFROM ProductWHERE Price > 100
SELECT Pname AS prodName, Price AS askPriceFROM ProductWHERE Price > 100
Product
prodName askPrice
SingleTouch $149.99
MultiTouch $203.99Query withrenaming
![Page 8: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/8.jpg)
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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)
![Page 9: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/9.jpg)
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(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))
![Page 10: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/10.jpg)
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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”)
![Page 11: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/11.jpg)
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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.productFROM PurchaseWHERE buyer = (SELECT name FROM Person WHERE ssn = ‘123456789‘);
SELECT Purchase.productFROM PurchaseWHERE buyer = (SELECT name FROM Person WHERE ssn = ‘123456789‘);
![Page 12: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/12.jpg)
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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 moreruntime errors
![Page 13: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/13.jpg)
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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 !
![Page 14: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/14.jpg)
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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
![Page 15: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/15.jpg)
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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’)
Nowthey are equivalent
![Page 16: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/16.jpg)
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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 producedBy “Gizmo-Works”
You can also use: s > ALL R s > ANY R EXISTS R
![Page 17: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/17.jpg)
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Question for Database Fansand 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
![Page 18: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/18.jpg)
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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...
![Page 19: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/19.jpg)
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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
![Page 20: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/20.jpg)
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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, makerFROM Product AS xWHERE price > ALL (SELECT price FROM Product AS y WHERE x.maker = y.maker AND y.year < 1972);
SELECT DISTINCT pname, makerFROM Product AS xWHERE price > ALL (SELECT price FROM Product AS y WHERE x.maker = y.maker AND y.year < 1972);
![Page 21: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/21.jpg)
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Aggregation
SELECT Avg(price)FROM ProductWHERE maker=“Toyota”
SELECT Avg(price)FROM ProductWHERE maker=“Toyota”
SQL supports several aggregation operations:
SUM, MIN, MAX, AVG, COUNT
![Page 22: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/22.jpg)
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Aggregation: Count
SELECT Count(*)FROM ProductWHERE year > 1995
SELECT Count(*)FROM ProductWHERE year > 1995
Except COUNT, all aggregations apply to a single attribute
![Page 23: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/23.jpg)
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Aggregation: Count
COUNT applies to duplicates, unless otherwise stated:
SELECT Count(category) same as Count(*)FROM ProductWHERE year > 1995
Better:
SELECT Count(DISTINCT category)FROM ProductWHERE year > 1995
![Page 24: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/24.jpg)
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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 PurchaseWHERE product = ‘bagel’
![Page 25: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/25.jpg)
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Simple Aggregations
Product Date Price Quantity
Bagel 10/21 0.85 15
Banana 10/22 0.52 7
Banana 10/19 0.52 17
Bagel 10/20 0.85 20
Purchase
![Page 26: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/26.jpg)
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Grouping and AggregationUsually, 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 TotalSalesFROM PurchaseWHERE date > “10/1”GROUPBY product
SELECT product, Sum(price*quantity) AS TotalSalesFROM PurchaseWHERE date > “10/1”GROUPBY product
Let’s see what this means…
![Page 27: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/27.jpg)
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Grouping and Aggregation
1. Compute the FROM and WHERE clauses.2. Group by the attributes in the GROUPBY3. Select one tuple for every group (and apply aggregation)
SELECT can have (1) grouped attributes or (2) aggregates.
![Page 28: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/28.jpg)
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First compute the FROM-WHERE clauses (date > “10/1”) then GROUP BY product:
Product Date Price Quantity
Banana 10/19 0.52 17
Banana 10/22 0.52 7
Bagel 10/20 0.85 20
Bagel 10/21 0.85 15
![Page 29: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/29.jpg)
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Then, aggregate
Product TotalSales
Bagel $29.75
Banana $12.48
SELECT product, Sum(price*quantity) AS TotalSalesFROM PurchaseWHERE date > “10/1”GROUPBY product
SELECT product, Sum(price*quantity) AS TotalSalesFROM PurchaseWHERE date > “10/1”GROUPBY product
![Page 30: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/30.jpg)
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GROUP BY v.s. Nested Queries
SELECT product, Sum(price*quantity) AS TotalSalesFROM PurchaseWHERE date > “10/1”GROUP BY product
SELECT product, Sum(price*quantity) AS TotalSalesFROM PurchaseWHERE 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 TotalSalesFROM Purchase xWHERE 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 TotalSalesFROM Purchase xWHERE x.date > “10/1”
![Page 31: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/31.jpg)
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Another Example
SELECT product, Sum(price * quantity) AS SumSales Max(quantity) AS MaxQuantityFROM PurchaseGROUP BY product
SELECT product, Sum(price * quantity) AS SumSales Max(quantity) AS MaxQuantityFROM PurchaseGROUP BY product
For every product, what is the total sales and max quantity sold?
Product SumSales MaxQuantity
Banana $12.48 17
Bagel $29.75 20
![Page 32: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/32.jpg)
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HAVING Clause
SELECT product, Sum(price * quantity)FROM PurchaseWHERE date > “9/1”GROUP BY productHAVING Sum(quantity) > 100
SELECT product, Sum(price * quantity)FROM PurchaseWHERE date > “9/1”GROUP BY productHAVING Sum(quantity) > 100
Same query, except that we consider only products that hadat least 100 buyers.
HAVING clause contains conditions on aggregates.
![Page 33: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/33.jpg)
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General form of Grouping and Aggregation
SELECT S
FROM R1,…,Rn
WHERE C1
GROUP BY a1,…,ak
HAVING C2
S = may contain attributes a1,…,ak and/or any aggregates but NO OTHER ATTRIBUTES
C1 = is any condition on the attributes in R1,…,Rn
C2 = is any condition on aggregate expressions
Why ?
![Page 34: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/34.jpg)
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General form of Grouping and Aggregation
SELECT S
FROM R1,…,Rn
WHERE C1
GROUP BY a1,…,ak
HAVING C2
Evaluation steps:1. Compute the FROM-WHERE part, obtain a table with all attributes
in R1,…,Rn
2. Group by the attributes a1,…,ak
3. Compute the aggregates in C2 and keep only groups satisfying C24. Compute aggregates in S and return the result
![Page 35: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/35.jpg)
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Aggregation
Author(login,name)
Document(url, title)
Wrote(login,url)
Mentions(url,word)
![Page 36: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/36.jpg)
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• Find all authors who wrote at least 10 documents:
• Attempt 1: with nested queries
SELECT DISTINCT Author.nameFROM AuthorWHERE count(SELECT Wrote.url FROM Wrote WHERE Author.login=Wrote.login) > 10
SELECT DISTINCT Author.nameFROM AuthorWHERE count(SELECT Wrote.url FROM Wrote WHERE Author.login=Wrote.login) > 10
This isSQL bya novice
![Page 37: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/37.jpg)
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• Find all authors who wrote at least 10 documents:
• Attempt 2: SQL style (with GROUP BY)
SELECT Author.nameFROM Author, WroteWHERE Author.login=Wrote.loginGROUP BY Author.nameHAVING count(wrote.url) > 10
SELECT Author.nameFROM Author, WroteWHERE Author.login=Wrote.loginGROUP BY Author.nameHAVING count(wrote.url) > 10
This isSQL byan expert
No need for DISTINCT: automatically from GROUP BY
![Page 38: 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:](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e495503460f94b3c7ab/html5/thumbnails/38.jpg)
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• Find all authors who have a vocabulary over 10000 words:
SELECT Author.nameFROM Author, Wrote, MentionsWHERE Author.login=Wrote.login AND Wrote.url=Mentions.urlGROUP BY Author.nameHAVING count(distinct Mentions.word) > 10000
SELECT Author.nameFROM Author, Wrote, MentionsWHERE Author.login=Wrote.login AND Wrote.url=Mentions.urlGROUP BY Author.nameHAVING count(distinct Mentions.word) > 10000
Look carefully at the last two queries: you maybe tempted to write them as a nested queries,but in SQL we write them best with GROUP BY