© פרופ' יהושע שגיב, האוניברסיטה העברית SQL © פרופ' יהושע שגיב, האוניברסיטה העברית

Basic SQL Query SELECT [Distinct] target-list FROM from-list WHERE condition; from-list: A list of relation names (possibly with a range-variable after each name) target-list: A list of fields of relations in relation-list condition: A boolean condition DISTINCT: Optional keyword to delete duplicates

Basic SQL Query (continues) SELECT [Distinct] target-list FROM from-list WHERE condition; The result evaluation: Compute the cross product of the tables in from-list. Delete all rows that do not satisfy condition. Delete all columns that do not appear in target-list. If Distinct is specified eliminate duplicate rows.

Basic SQL Query (continues) SELECT Distinct A1,…,An FROM R1,…,Rm WHERE C; This translates to the expression in relational algebra: A1,…,An (C(R1 x…x Rm))

Example Tables Used Sailors sid sname rating age 22 31 58 Dustin Lubber Rusty 7 8 10 45.0 55.5 35.0 Boats bid bname color 101 103 Nancy Gloria red green Reserves sid bid Day 22 58 101 103 10/10/02 11/12/02 Key Key

Boat Names and Reservation Dates SELECT bname,day FROM Boats,Reserves WHERE Boats.bid = Reserves.bid

Boat Names and Reservation Dates SELECT bname,day FROM Boats,Reserves WHERE Boats.bid = Reserves.bid Reserves sid bid day 22 101 10/10/02 58 103 11/12/02 Boats bid bname color 101 103 Nancy Gloria red green

Boat Names and Reservation Dates SELECT bname,day FROM Boats,Reserves WHERE Boats.bid = Reserves.bid Reserves sid bid day 22 101 10/10/02 58 103 11/12/02 Boats bid bname color 101 Nancy red 103 Gloria green

Boat Names and Reservation Dates SELECT bname,day FROM Boats,Reserves WHERE Boats.bid = Reserves.bid Reserves day 10/10/02 11/12/02 Boats bname Nancy Gloria

Sailors Who Reserved Boat 103 SELECT sname FROM Sailors, Reserves WHERE Sailors.sid = Reserves.sid and bid = 103; sname(Sailors.sid = Reserves.sid  bid = 103 (Sailors x Reserves))

Sailors Who Reserved Boat 103 Sailors x Reserves Sailors Reserves sid sname rating age bid day 22 Dustin 7 45.0 101 10/10/02 58 103 11/12/02 31 Lubber 8 55.5 Rusty 10 35.0

Sailors Who Reserved Boat 103 Sailors.sid = Reserves.sid  bid = 103 Sailors Reserves sid sname rating age bid day 22 Dustin 7 45.0 101 10/10/02 58 103 11/12/02 31 Lubber 8 55.5 Rusty 10 35.0

Sailors Who Reserved Boat 103 sname Sailors Reserves sid sname rating age bid day 22 Dustin 7 45.0 101 10/10/02 58 103 11/12/02 31 Lubber 8 55.5 Rusty 10 35.0

Range Variables Range variables are good style SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid = R.sid and R.bid = 103; Range variables are good style They are necessary if the same relation appears twice in the FROM clause

A Few SELECT Options Rename selected columns: Select all columns: FROM Sailors S; Rename selected columns: SELECT S.sname AS Sailors_Name

A Few SELECT Options Applying functions: Mathematical manipulations: SELECT (age-5)*2 FROM Sailors S; Aggregate functions: SELECT COUNT(*) SELECT MAX(age)

The WHERE Clause Numerical and string comparison: !=,<>,=, <, >, >=, <=, between(between val1 AND val2) String comparisson is according to the alphabetical order! Logical components: AND, OR Null verification: IS NULL, IS NOT NULL Example: SELECT sname FROM Sailors WHERE age>=40 AND rating IS NOT NULL ;

The WHERE Clause (continues) The LIKE operator: A pattern matching operator Basic format: colname LIKE pattern Example: _ is a single character % is 0 or more characters SELECT sid FROM Sailors WHERE sname LIKE ‘B_%g’;

Sailors Who’ve Reserved a Boat SELECT sname FROM Sailors S, Reserves R WHERE S.sid = R.sid; Would adding DISTINCT give a different result?

Exercise Formulate a query that finds the names of sailors who reserved a yellow boat. Sailors sid sname rating age 22 31 58 Dustin Lubber Rusty 7 8 10 45.0 55.5 35.0 Boats bid bname color 101 103 Nancy Gloria red green Reserves sid bid Day 22 58 101 103 10/10/02 11/12/02

A Harder Exercise Formulate a query that finds the bid of boats that are reserved in at least two different days Fix the query you formulated to find the names of these boats Hint: A relation may appear more than once in the FROM list…

Union, Intersect and Except Sqlplus supports the union, intersection and difference operators The syntax: query1 UNION query2; query1 INTERSECT query2; query1 MINUS query2;

Sailors who’ve reserved a red or green boat SELECT S.sname FROM Sailors S, Boats B, Reserves R WHERE S.sid = R.sid and R.bid = B.bid and (B.color = ‘red’ or B.color = ‘green’); sname(color = ‘red’  color = ‘green’ (Sailors ⋈ Reserves ⋈ Boats)) What would happen if we replaced or by and ?

Sailors who’ve reserved red or green boat SELECT S.sname FROM Sailors S, Boats B, Reserves R WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘red’ UNION WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘green’;

The Second Version in Relational Algebra sname(color = ‘red’ (Sailors ⋈ Reserves ⋈ Boats))  sname(color = ‘green’ (Sailors ⋈ Reserves ⋈ Boats))

Nested Queries A query may be nested within another through the following operators: (NOT) IN (NOT) EXISTS ANY ALL A query with nested queries is computed using nested loops

The IN Operator Names of sailors who’ve reserved boat 103: SELECT S.sname FROM Sailors S WHERE S.sid IN (SELECT R.sid FROM Reserves R WHERE R.bid = 103); What would happen if we wrote NOT IN?

Notice the correlation between the examined row and the inner query The EXISTS Operator Names of sailors who’ve reserved boat 103: SELECT S.sname FROM Sailors S WHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid = 103 and S.sid = R.sid); What would happen if we wrote NOT EXISTS? Notice the correlation between the examined row and the inner query

Set-Comparison Queries: ANY,ALL Sailors who are not the youngest: SELECT * FROM Sailors S1 WHERE S1.age > ANY (SELECT S2.age FROM Sailors S2); We can also use op ALL (op is >, <, =, >=, <=, or <>)

Another Exercise What does the following query compute? Using IN and INTERSECT operators, formulate a query that finds the names of the sailors who reserved both green and red boats. SELECT S.sname FROM Sailors S, Boats B, Reserves R WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘red’ INTERSECT WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘green’;

Another Exercise (continues) What does the following query compute? SELECT S.sname FROM Sailors S WHERE NOT EXISTS ( (SELECT B.bid FROM Boats B) MINUS (SELECT R.bid FROM Reserves R WHERE R.sid=S.sid) ) ;

For The Brave Ones… Formulate a query that computes the sid of all sailors who have not reserved a green boat. Prove formally that in the Sailors-Reserves-Boats database, the query of question 1 cannot be as simple as a query of the form: SELECT Vj.sid FROM R1 V1, R2 V2…Rn Vn WHERE C Where C is a simple condition (contains only comparison clauses, AND and OR, and does not contain nested queries) Notice that the records in each table are arbitrary!