1 SQL: Structured Query Language (‘Sequel’) Chapter 5.

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SQL: Structured Query Language(‘Sequel’)

Chapter 5

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Running Example

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

sid bid day

22 101 10/10/9658 103 11/12/96

R1

S1

S2

Instances of the Sailors and Reserves relations in our examples.

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Basic SQL Query

relation-list A list of relation names target-list A list of attributes of relations in

relation-list qualification

• Comparisons (Attr op const or Attr1 op Attr2, where op is one of ) combined using AND, OR and NOT.

DISTINCT is optional keyword indicating that answer should not contain duplicates. Default is that duplicates are not eliminated!

SELECT [DISTINCT] target-listFROM relation-listWHERE qualification

, , , , ,

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Example Query

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103

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Comparisons in Oracle

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Conceptual Evaluation Strategy

Semantics of SQL query defined in terms of conceptual evaluation strategy:

• Compute cross-product of relation-list.• Discard resulting tuples if they fail

qualifications.• Delete attributes that are not in target-list.• If DISTINCT is specified, eliminate duplicate

rows. Probably the least efficient way! Optimizer must find more efficient strategies

to compute same answers.


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22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid bid day

22 101 10/10/9658 103 11/12/96

R S

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 22 101 10/ 10/ 96

22 dustin 7 45.0 58 103 11/ 12/ 96

31 lubber 8 55.5 22 101 10/ 10/ 96

31 lubber 8 55.5 58 103 11/ 12/ 96

58 rusty 10 35.0 22 101 10/ 10/ 96

58 rusty 10 35.0 58 103 11/ 12/ 96


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A Note on Range Variables Needed only if same relation appears twice in FROM

clause. The previous query can also be written as:

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid AND bid=103

SELECT snameFROM Sailors, Reserves WHERE Sailors.sid=Reserves.sid AND bid=103

It is good style, however, to use range variables always!

OR

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Find sailors who’ve reserved at least one boat

Question: What is the output of this query ?

SELECT S.sidFROM Sailors S, Reserves RWHERE S.sid=R.sid

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Expressions and Strings

AS and = are two ways to name fields in result. LIKE is used for string matching. `_’ stands for any one

character and `%’ stands for 0 or more arbitrary characters.

MEANING of Query: Find sailors (age of sailor and two fields defined by

expressions) whose names begin and end with B and contain at least three characters.

SELECT S.age, age1=S.age-5, 2*S.age AS age2FROM Sailors SWHERE S.sname LIKE ‘B_%B’

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SQL Queries Using Set Operations

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UNIONcompute union of any two union-compatible sets of tuples

SELECT S.sidFROM Sailors S, Reserves R, Boats BWHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ OR B.color=‘green’)

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’UNIONSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red or a green boat

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If we replace OR by AND as done below, what are the semantics?

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ AND B.color=‘green’)

Find sid’s of sailors who’ve reserved a red and a green boat.

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SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’

INTERSECT

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red and a green boat

INTERSECT: compute intersection of two union-compatible sets of tuples.

What happens if change S.sid as S.sname ?

Key field!

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SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’INTERSECTSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red and a green boat

Could query be written without INTERSECT?

SELECT S.sidFROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2WHERE (S.sid=R1.sid AND R1.bid=B1.bid AND B1.color=‘red’ ) AND (S.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

Do we need Sailors relation in the FROM clauses?

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SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’EXCEPTSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red and not a green boat

What if we want to compute the DIFFERENCE instead?

SELECT S.sidFROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color='red' AND NOT(B2.color = 'green'))

SQL/92 standard, but some systems don’t support EXCEPT.

What then?

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SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’EXCEPTSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red and not a green boat

What if we want to compute the DIFFERENCE instead?

SELECT S.sidFROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color='red' AND NOT(B2.color = 'green'))

Except is SQL/92 standard, but some systems don’t . support EXCEPT.

Is the above query correct, or not correct ?

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About duplication

Default: Eliminate duplicate

Unless we use special keyword “ALL” :

# of copies of a row in result

UNION ALL (m+n)INTERSECT ALL min(m,n)EXCEPT ALL m-n

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Nested Queries

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Nested Queries

WHERE, FROM, HAVING clauses can itself contain another SQL query, called a subquery.

powerful feature of SQL.

SELECT S.snameFROM Sailors SWHERE S.sid IN (SELECT R.sid FROM Reserves R WHERE R.bid=103)

Find names of sailors who’ve reserved boat #103:

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Nested Queries Evaluation

Nested loops evaluation: For each Sailors tuple,

• Evaluate the WHERE clause qualification• Which here means re-compute subquery.



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Nested Queries

Change query to find sailors who’ve not reserved #103:



SELECT S.snameFROM Sailors SWHERE S.sid NOT IN (SELECT R.sid FROM Reserves R WHERE R.bid=103)

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Nested Queries

Could this query be rewritten without nesting?



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Nested Queries with Correlation

EXISTS is another set comparison operator, like IN.

SELECT S.snameFROM Sailors SWHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid)


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Nested Queries with Correlation

Why, in general, this subquery must be re-computed for each Sailors tuple ?

Could this query be rewritten into a flat SQL query?

SELECT S.snameFROM Sailors SWHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid)


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More on Set-Comparison Operators op ANY and op ALL where op = { }

Find sailors whose rating is greater than that of some sailor called Horatio:

IN is equal to =ANY

NOT IN is equal to <>ANY

, , , , ,

SELECT *FROM Sailors SWHERE S.rating > ANY (SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)

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More on Set-Comparison Operators

What if there is no other sailor with that name? “> ALL” would return TRUE if set is empty.

SELECT *FROM Sailors SWHERE S.rating > ALL (SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)

Find sailors whose rating is greater than that of all sailors called Horatio:

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Rewriting INTERSECT Queries Using IN

Find sid’s of sailors who’ve reserved both a red and a green boat:

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN (SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

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Rewriting INTERSECT Queries Using IN

INTERSECT queries re-written using IN.

Note: This query can return “sname” (and not just “sid) of Sailors who’ve reserved both red and green boats

Find sname’s of sailors who’ve reserved both a red and a green boat:

SELECT S.snameFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN (SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

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Division in SQLFind sailors who’ve reserved all boats.

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Division in SQL

SELECT S.snameFROM Sailors SWHERE NOT EXISTS ((SELECT B.bid FROM Boats B) EXCEPT (SELECT R.bid FROM Reserves R WHERE R.sid=S.sid))

Find sailors who’ve reserved all boats.

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Division in SQL

Can we do it without EXCEPT ?



(1)

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Division in SQL

Let’s do it without EXCEPT:


SELECT S.snameFROM Sailors SWHERE NOT EXISTS (SELECT B.bid FROM Boats B WHERE NOT EXISTS (SELECT R.bid FROM Reserves R WHERE R.bid=B.bid AND R.sid=S.sid))

Sailors S such that ...

there is no boat B without ...

a Reserves tuple showing S reserved B


(1)

(2)

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Aggregation and Having Clauses

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Aggregate Operators

Significant extension of relational algebra.

COUNT (*)COUNT ( [DISTINCT] A)SUM ( [DISTINCT] A)AVG ( [DISTINCT] A)MAX (A)MIN (A)

Why no Distinct?

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Aggregate Operators

COUNT (*)COUNT ( [DISTINCT] A)SUM ( [DISTINCT] A)AVG ( [DISTINCT] A)MAX (A)MIN (A)

SELECT AVG (S.age)FROM Sailors SWHERE S.rating=10

SELECT COUNT (*)FROM Sailors S

Why no Distinct?

SELECT COUNT (DISTINCT S.rating)FROM Sailors SWHERE S.sname=‘Bob’

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More Examples of Aggregate Operators

SELECT AVG ( DISTINCT S.age)FROM Sailors SWHERE S.rating=10

SELECT S.snameFROM Sailors SWHERE S.rating= (SELECT MAX(S2.rating) FROM Sailors S2)

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Find name and age of the oldest sailor(s)

SELECT S.sname, MAX (S.age)FROM Sailors S

SELECT S.sname, S.ageFROM Sailors SWHERE S.age = (SELECT MAX (S2.age) FROM Sailors S2)

•Is this first query legal?

Is this second query legal?

•No! •Why not ?

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Motivation for Grouping So far, aggregate operators to all (qualifying)

tuples. Question?

What if want to apply aggregate to each group of tuples. Example :

Find the age of the youngest sailor for each rating level. Example Procedure :

Suppose rating values {1,2,…,10}, 10 queries:

SELECT MIN (S.age)FROM Sailors SWHERE S.rating = i

For i = 1, 2, ... , 10:

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Motivation for Grouping

What’s the problem with above ? • We don’t know how many rating levels

exist.• Nor what the rating values for these

levels are.

SELECT MIN (S.age)FROM Sailors SWHERE S.rating = i

For i = 1, 2, ... , 10:

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Add Group By to SQL

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Queries With GROUP BY and HAVING

A group is a set of tuples that each have the same value for all attributes in grouping-list.

HAVING clause is a restriction on each group.

SELECT [DISTINCT] target-listFROM relation-listWHERE qualificationGROUP BY grouping-listHAVING group-qualification

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Queries With GROUP BY and HAVING

target-list contains :(i) attribute names (ii) aggregate-op (column-name) (e.g., MIN (S.age)).

A group is a set of tuples that each have the same value for all attributes in grouping-list.

REQUIREMENT: - target-list (i) grouping-list. - Why?

- Each answer tuple of a group must have single value.

SELECT [DISTINCT] target-listFROM relation-listWHERE qualificationGROUP BY grouping-listHAVING group-qualification

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Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors

SELECT S.rating, MIN (S.age) AS min-ageFROM Sailors SWHERE S.age >= 18GROUP BY S.ratingHAVING COUNT (*) > 1

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GroupBy --- Conceptual Evaluation

1. Compute the cross-product of relation-list2. Discard tuples that fail qualification 3. Delete `unnecessary’ fields4. Partition the remaining tuples into groups by the value

of attributes in grouping-list. (GroupBy)5. Eliminate groups using the group-qualification (Having)

We will have a single value per group That is, one answer tuple is generated per qualifying

group.

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Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors

SELECT S.rating, MIN (S.age) AS min-age

FROM Sailors SWHERE S.age >= 18GROUP BY S.ratingHAVING COUNT (*) > 1


22 dustin 7 45.0

29 brutus 1 33.0

31 lubber 8 55.5

32 andy 8 25.5

58 rusty 10 35.0

64 horatio 7 35.0

71 zorba 10 16.0

74 horatio 9 35.0

85 art 3 25.5

95 bob 3 63.5

96 frodo 3 25.5

Sailors instance:

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Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.

rating age

7 45.0

1 33.0

8 55.5

8 25.5

10 35.0

7 35.0

10 16.0

9 35.0

3 25.5

3 63.5

3 25.5

rating minage 3 25.5 7 35.0 8 25.5

rating age

1 33.0

3 25.5

3 63.5

3 25.5

7 45.0

7 35.0

8 55.5

8 25.5

9 35.0

10 35.0

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Find age of the youngest sailor with age 18, for each rating with at least 2 sailors between 18 and 60.

SELECT S.rating, MIN (S.age) AS min-age

FROM Sailors SWHERE S.age >= 18 AND S.age <= 60GROUP BY S.ratingHAVING COUNT (*) > 1


22 dustin 7 45.0

29 brutus 1 33.0

31 lubber 8 55.5

32 andy 8 25.5

58 rusty 10 35.0

64 horatio 7 35.0

71 zorba 10 16.0

74 horatio 9 35.0

85 art 3 25.5

95 bob 3 63.5

96 frodo 3 25.5

Answer relation:

Sailors instance:

rating minage 3 25.5 7 35.0 8 25.5

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For each red boat, find the number of reservations for this boat

Grouping over Join of three relations. Q: What do we get if we remove B.color=‘red’ from

WHERE clause and add HAVING clause with this condition?

• No difference. But Illegal.• Only column in GroupBy can appear in Having,

unless in aggregate operator of Having

SELECT B.bid, COUNT (*) AS s-countFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’GROUP BY B.bid

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Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age)

HAVING COUNT(*) >1

SELECT S.rating, MIN (S.age)FROM Sailors SWHERE S.age > 18GROUP BY S.ratingHAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating)

HAVING clause can also contain a subquery.

Find age of the youngest sailor with age > 18, for each rating with at least 2 such sailors (of age 18)

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Find those ratings for which the average age is the minimum over all ratings

WRONG : Aggregate operations cannot be nested!

SELECT S.ratingFROM Sailors SWHERE S.age = (SELECT MIN (AVG (S2.age))

FROM Sailors S2)

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SELECT Temp.rating, Temp.avg-age

FROM (SELECT S.rating, AVG (S.age) AS avg-age FROM Sailors S GROUP BY S.rating) AS Temp

WHERE Temp.avg-age = (SELECT MIN (avg-age) FROM (SELECT S.rating, AVG(S.age) as avg_age FROM Sailors S GROUP BY S.rating))

Find those ratings for which the average age is the minimum over all ratings

Correct solution (in SQL/92):

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Null Values

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Null Values

Field values in a tuple are sometimes : Unknown (e.g., a rating has not been assigned) or Inapplicable (e.g., no maiden-name when a male person)

SQL provides a special value null for such situations.

The presence of null complicates many issues.

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Comparisons Using Null Values

(S.rating = 8) is TRUE or FALSE ?? What if rating has null value in tuple ?

We need a 3-valued logic : condition can be true, false or unknown.

SQL special operators IS NULL to check if value is/is not null.

Disallow NULL value : rating INTEGER NOT NULL

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Comparison with NULL values

Arithmetic operations on NULL return NULL.

Comparison operators on NULL return UNKNOWN.

We can explicitly check whether a value is null or not by IS NULL and by IS NOT NULL.

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Truth table with UNKNOWN

UNKNOWN AND TRUE = UNKNOWNUNKNOWN OR TRUE = TRUEUNKNOWN AND FALSE = FALSEUNKNOWN OR FALSE = UNKNOWNUNKNOWN AND UNKNOWN = UNKNOWNUNKNOWN OR UNKNOWN = UNKNOWNNOT UNKNOWN = UNKNOWN

WHERE clause is satisfied only when it evaluates to TRUE.

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Outer Joins : Special Operators

Left Outer Join Right Outer Join Full Outer Join

SELECT S.sid, R.bidFROM Sailors S NATURAL LEFT OUTER JOIN Reserves R

SELECT S.sid, R.bidFROM Sailors S LEFT OUTER JOIN Reserves RON S.sid = R.sid

Sailors rows (left) without a matching Reserves row (right) appear in result, but not vice versa.

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Sorting: ORDER BY clause

SELECT * FROM StudentWHERE sNumber >= 1ORDER BY sNumber, sName

(sNumber, sName) ( (sNumber >= 1) (Student))

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SQL Queries - Summary

SELECT [DISTINCT] a1, a2, …, anFROM R1, R2, …, Rm[WHERE C1][GROUP BY g1, g2, …, gl [HAVING C2]][ORDER BY o1, o2, …, oj]

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Summary SQL was important factor in early acceptance of

relational model : easy-to-understand !

Relationally complete: in fact, significantly more expressive power than relational algebra.

Many alternative ways to write a query. So optimizer must look for most efficient evaluation plan.

In practice, users (still) need to be aware of how queries are optimized and evaluated for best results.

1 SQL: Structured Query Language (‘Sequel’) Chapter 5.

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