SQL: Queries, Constraints, Triggers

SQL: Queries, Constraints, Triggers

SQL: Queries, Constraints, Triggers Chapter 5 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke DML versus DDL Data Manipulation Language (DML) posing queries and operating on tuples Data Definition Language (DDL) operating on tables/views atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke sid bid day Instances 22 101 10/10/96 58 103 11/12/96 R1 Example We will use these S1

instances of the Sailors and Reserves relations in our examples. If the key for the Reserves relation S2 contained only the attributes sid and bid, how would the semantics differ? sid 22 31 58 sname rating age dustin 7 45.0 lubber 8 55.5 rusty 10 35.0 sid 28 31 44 58

sname rating age yuppy 9 35.0 lubber 8 55.5 guppy 5 35.0 rusty 10 35.0 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke Basic SQL Query SELECT [DISTINCT] target-list FROM WHERE relation-list qualification relation-list A list of relation names (possibly with a range-variable after each name). 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 an optional keyword indicating that the answer should not contain duplicates. Default is that duplicates are not eliminated! atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke Conceptual Evaluation Strategy Semantics of an SQL query defined in terms of the following conceptual evaluation strategy: Compute the 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. This strategy is probably the least efficient way to compute a query! An optimizer will find

more efficient strategies to compute the same answers. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke Example of Conceptual Evaluation SELECT FROM WHERE S.sname Sailors S, Reserves R S.sid=R.sid AND R.bid=103 (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 31 lubber 8 8 55.5 55.5 22 58 101 10/ 10/ 96 103 11/ 12/ 96 58 rusty 10 35.0 22 101 10/ 10/ 96 58 rusty 10

35.0 58 103 11/ 12/ 96 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke A Note on Range Variables Really needed only if the same relation appears twice in the FROM clause. The previous query can also be written as: SELECT FROM WHERE OR S.sname Sailors S, Reserves R S.sid=R.sid AND bid=103 SELECT FROM WHERE sname Sailors, Reserves

Sailors.sid=Reserves.sid AND bid=103 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke It is good style, however, to use range variables always! Find sailors whove reserved at least one boat SELECT FROM S.sid Sailors S, Reserves R WHERE S.sid=R.sid Would adding DISTINCT to this query make a difference? What is the effect of replacing S.sid by S.sname in the SELECT clause? Would adding DISTINCT to this variant of the query make a difference? atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke Expressions and Strings

SELECT S.age, age1=S.age-5, FROM Sailors S WHERE S.sname LIKE B_%B 2*S.age AS age2 Illustrates use of arithmetic expressions and string pattern matching: Find triples (of ages of sailors and two fields defined by expressions) for sailors whose names begin and end with B and contain at least three characters. 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. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke Find sids of sailors whove reserved a red or a green boat UNION:

Can be used to compute the union of any two unioncompatible sets of tuples (which are themselves the result of SQL queries). If we replace OR by AND in the first version, what do we get? Also available: EXCEPT (What do we get if we replace UNION by EXCEPT?) SELECT S.sid 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 SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=red UNION SELECT S.sid FROM Sailors S, Boats B, Reserves R atabase Management Systems 3ed, R. Ramakrishnan

and J. Gehrke 1 Find sids of sailors whove reserved a red and a green boat SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, INTERSECT: Can be used Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid to compute the intersection of any two AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color=red AND B2.color=green union-compatible sets of tuples. Key field! SELECT S.sid Included in the SQL/92 FROM Sailors S, Boats B, Reserves R standard, but some systems dont support it. WHERE S.sid=R.sid AND

R.bid=B.bid Contrast symmetry of AND B.color=red the UNION and INTERSECT INTERSECT queries with how much SELECT S.sid the other versions differ. FROM Sailors S, Boats B, atabase Management Systems 3ed, Reserves R WHERE S.sid=R.sid AND R. Ramakrishnan and J. Gehrke 1 Nested Queries Find names of sailors whove reserved boat #103: SELECT S.sname FROM Sailors S WHERE S.sid IN (SELECT R.sid FROM Reserves R WHERE R.bid=103) A very powerful feature of SQL: a WHERE clause can itself contain an SQL query! (Actually, so can FROM and HAVING clauses.) To find sailors whove not reserved #103, use NOT IN.

To understand semantics of nested queries, think of a nested loops evaluation: For each Sailors tuple, check the qualification by computing the subquery. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Nested Queries with Correlation Find names of sailors whove reserved boat #103: SELECT S.sname FROM Sailors S WHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid) EXISTS is another set comparison operator, like IN. Illustrates why, in general, subquery must be recomputed for each Sailors tuple. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1

More on Set-Comparison Operators Weve already seen IN, EXISTS and UNIQUE. Can also use NOT IN, NOT EXISTS and NOT UNIQUE. Also available: op ANY, op ALL, opIN , , , , , Find sailors whose rating is greater than that of some sailor called Horatio: SELECT * FROM Sailors S WHERE S.rating > ANY (SELECT S2.rating FROM Sailors S2 WHERE S2.sname=Horatio) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Rewriting INTERSECT Queries Using IN Find sids of sailors whove reserved both a red and a green boat: LECT S.sid OM Sailors S, Boats B, Reserves R HERE 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 R WHERE S2.sid=R2.sid AND R2.bid=B2.b AND B2.color=green) Similarly, EXCEPT queries re-written using NOT IN. To find names (not sids) of Sailors whove reserved both red and green boats, just replace S.sid by S.sname in SELECT clause. (What about INTERSECT query?) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 (1)SELECT S.sname Division in SQL ind sailors whove reserved all Lets do it the hard way, without EXCEPT: LECT (2) S.sname OM Sailors S HERE NOT EXISTS FROM Sailors S WHERE NOT EXISTS

((SELECT B.bid FROM Boats B) EXCEPT boats. (SELECT R.bid FROM Reserves R WHERE R.sid=S.sid)) (SELECT B.bid FROM Boats B Sailors S such that ... WHERE NOT EXISTS (SELECT R.bid FROM Reserves R WHERE R.bid=B. there is no boat B without ... AND R.sid=S a Reserves tuple showing S reserved B atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 COUNT (*) Aggregate Operators Significant extension relational algebra. SELECT COUNT (*) FROM Sailors S

COUNT ( [DISTINCT] A) SUM ( [DISTINCT] A) AVG ( [DISTINCT] A) ofMAX (A) MIN (A) single column SELECT S.sname FROM Sailors S WHERE S.rating= (SELECT MAX(S2.rating) FROM Sailors S2) SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10 SELECT COUNT (DISTINCT FROM Sailors S WHERE S.sname=Bob S.rating) SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Find name and age of the oldest sailor(s) The first query is illegal!

(Well look into the reason a bit later, when we discuss GROUP BY.) The third query is equivalent to the second query, and is allowed in the SQL/92 standard, but is not supported in some systems. SELECT S.sname, MAX FROM Sailors S (S.age) SELECT S.sname, S.age FROM Sailors S WHERE S.age = (SELECT MAX (S2.age) FROM Sailors S2) SELECT S.sname, S.age FROM Sailors S WHERE (SELECT MAX (S2.age) FROM Sailors S2) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke = S.age 1

Motivation for Grouping So far, weve applied aggregate operators to all (qualifying) tuples. Sometimes, we want to apply them to each of several groups of tuples. Consider: Find the age of the youngest sailor for each rating level. In general, we dont know how many rating levels exist, and what the rating values for these levels are! Suppose we know that rating values go from 1 to 10; we can write 10 queries that look like this (!): For i = 1, 2, ... , 10: SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Queries With GROUP BY and HAVING SELECT

[DISTINCT] target-list relation-list qualification grouping-list groupqualification FROM WHERE GROUP BY HAVING The target-list contains (i) attribute names (ii) terms with aggregate operations (e.g., MIN (S.age)). The attribute list (i) must be a subset of grouping-list. Intuitively, each answer tuple corresponds to a group, and these attributes must have a single value per group. (A group is a set of tuples that have the same value for all attributes in grouping-list.) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Conceptual Evaluation The cross-product of relation-list is computed, tuples that fail qualification are discarded, `unnecessary fields are deleted, and the remaining tuples are partitioned into groups by the value of attributes in

grouping-list. The group-qualification is then applied to eliminate some groups. Expressions in group-qualification must have a single value per group! In effect, an attribute in group-qualification that is not an argument of an aggregate op also appears in grouping-list. (SQL does not exploit primary key semantics here!) One answer tuple is generated per qualifying group. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 SELECT

Answer relation: rating 3 7 8 minage 25.5 35.0 25.5 Sailors instance: sid 22 29 31 32 58 64 71 74 85 95 96 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke sname rating age dustin 7 45.0 brutus

1 33.0 lubber 8 55.5 andy 8 25.5 rusty 10 35.0 horatio 7 35.0 zorba 10 16.0 horatio 9 35.0 art 3 25.5 bob 3 63.5 frodo 3 25.5 2 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors. rating 7 1 8 8 10

7 10 9 3 3 3 age 45.0 33.0 55.5 25.5 35.0 35.0 16.0 35.0 25.5 63.5 25.5 rating 1 3 3 3 7 7 8 8 9 10

age 33.0 25.5 63.5 25.5 45.0 35.0 55.5 25.5 35.0 35.0 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke rating 3 7 8 minage 25.5 35.0 25.5 2 Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors and with every sailor under 60. HAVING COUNT (*) > 1 AND EVERY (S.age <=60) rating

7 1 8 8 10 7 10 9 3 3 3 age 45.0 33.0 55.5 25.5 35.0 35.0 16.0 35.0 25.5 63.5 25.5 rating 1 3 3 3 7 7

8 8 9 10 age 33.0 25.5 63.5 25.5 45.0 35.0 55.5 25.5 35.0 35.0 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke rating minage 7 35.0 8 25.5 What is the result of changing EVERY to ANY? 2 Find age of the youngest sailor with

age 18, for each rating with at least 2 sailors between 18 and 60. Sailors instance: S.rating, MIN (S.age) sid sname rating age AS minage 22 dustin 7 45.0 FROM Sailors S 29 brutus 1 33.0 WHERE S.age >= 18 AND S.age <= 60 31 lubber 8 55.5 GROUP BY S.rating 32 andy 8 25.5 HAVING COUNT (*) > 1 SELECT Answer relation: rating 3 7 8 minage 25.5 35.0 25.5

58 64 71 74 85 95 96 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke rusty horatio zorba horatio art bob frodo 10 7 10 9 3 3 3 35.0 35.0 16.0 35.0 25.5

63.5 25.5 2 For each red boat, find the number of reservations for this boat SELECT B.bid, COUNT (*) AS scount FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND GROUP BY B.bid B.color=red Grouping over a join of three relations. What do we get if we remove B.color=red from the WHERE clause and add a HAVING clause with this condition? What if we drop Sailors and the condition involving S.sid? atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age) SELECT S.rating, MIN (S.age)

FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating) Shows HAVING clause can also contain a subquery. Compare this with the query where we considered only ratings with 2 sailors over 18! What if HAVING clause is replaced by: HAVING COUNT(*) >1 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Find those ratings for which the average age is the minimum over all ratings Aggregate operations cannot be nested! WRONG: SELECT S.rating FROM Sailors S WHERE S.age =

(SELECT MIN (AVG (S2.age)) FROM Sailors S2) Correct solution (in SQL/92): SELECT Temp.rating, Temp.avgage FROM (SELECT S.rating, AVG (S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avgage = (SELECT MIN (Temp.avgage) FROM Temp) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Null Values Field values in a tuple are sometimes unknown (e.g., a rating has not been assigned) or inapplicable (e.g., no spouses name). SQL provides a special value null for such situations. The presence of null complicates many issues. E.g.:

Special operators needed to check if value is/is not null. Is rating>8 true or false when rating is equal to null? What about AND, OR and NOT connectives? We need a 3-valued logic (true, false and unknown). Meaning of constructs must be defined carefully. (e.g., WHERE clause eliminates rows that dont evaluate to true.) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Adding and Deleting Tuples Can insert a single tuple using: INSERT INTO Students (sid, name, login, age, gpa) VALUES (53688, Smith, [email protected], 18, 3.2) Can delete all tuples satisfying some condition (e.g., name = Smith): DELETE FROM Students WHERE S.name S = Smith

Powerful variants of these commands are available; more la atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Updating Tuples atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Creating Relations in SQL Creates the Students CREATE TABLE Students (sid: CHAR(20), relation. Observe that the name: CHAR(20), type (domain) of each field is specified, and enforced by login: CHAR(10), the DBMS whenever tuples age: INTEGER, gpa: REAL) are added or modified. As another example, the CREATE TABLE Enrolled Enrolled table holds (sid: CHAR(20), information about courses cid: CHAR(20), that students take. grade:

CHAR(2)) atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Destroying and Altering Relations DROP TABLE Students Destroys the relation Students. The schema information and the tuples are deleted. ALTER TABLE Students ADD COLUMN firstYear: integer The schema of Students is altered by adding a new field; every tuple in the current instance is extended with a null value Systems in the field.and J. Gehrke atabase Management 3ed,new R. Ramakrishnan

3 Views A view is just a relation, but we store a definition, rather than a set of tuples. CREATE VIEW YoungActiveStudents (name, AS SELECT S.name, E.grade FROM Students S, Enrolled E WHERE S.sid = E.sid and S.age<21 Views can be dropped using the grade) DROP VIEW command. How to handle DROP TABLE if theres a view on the table? DROP TABLE command has options to let the user atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Questions atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke

3 Summary SQL was an important factor in the early acceptance of the relational model; more natural than earlier, procedural query languages. Relationally complete; in fact, significantly more expressive power than relational algebra. Even queries that can be expressed in RA can often be expressed more naturally in SQL. Many alternative ways to write a query; optimizer should look for most efficient evaluation plan. In practice, users need to be aware of how queries are optimized and evaluated for best results. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Summary (Contd.) NULL for unknown field values brings many complications SQL allows specification of rich integrity constraints Triggers respond to changes in the database

atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 Integrity Constraints (Review) An IC describes conditions that every legal instance of a relation must satisfy. Inserts/deletes/updates that violate ICs are disallowed. Can be used to ensure application semantics (e.g., sid is a key), or prevent inconsistencies (e.g., sname has to be a string, age must be < 200) Types of ICs: Domain constraints, primary key constraints, foreign key constraints, general constraints. Domain constraints: Field values must be of right type. Always enforced. atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke

3 CREATE TABLE Sailors ( sid INTEGER, sname CHAR(10), rating INTEGER, age REAL, Useful when PRIMARY KEY (sid), CHECK ( rating >= 1 more general AND rating <= 10 ICs than keys are involved. CREATE TABLE Reserves ( sname CHAR(10), Can use bid INTEGER, queries to day DATE, express PRIMARY KEY (bid,day), constraint. CONSTRAINT noInterlakeRes Constraints CHECK (`Interlake <> can be named. ( SELECT B.bname FROM Boats B WHERE B.bid=bid))) General Constraints

atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke ) 3 Constraints Over Multiple Relations CREATE TABLE Sailors ( sid INTEGER, Number of boats sname CHAR(10), plus number of Awkward and rating INTEGER, sailors is < 100 wrong! age REAL, If Sailors is PRIMARY KEY (sid), empty, the CHECK number of Boats tuples can be ( (SELECT COUNT (S.sid) FROM Sailors S) anything! + (SELECT COUNT (B.bid) FROM Boats B) < ASSERTION is the

10 right solution; not associated CREATE ASSERTION smallClub CHECK with either table. ( (SELECT COUNT (S.sid) FROM Sailors S) + (SELECT COUNT (B.bid) FROM Boats B) < 10 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 4 Triggers Trigger: procedure that starts automatically if specified changes occur to the DBMS Three parts: Event (activates the trigger) Condition (tests whether the triggers should run) Action (what happens if the trigger runs)

atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 4 Triggers: Example (SQL:1999) CREATE TRIGGER youngSailorUpdate AFTER INSERT ON SAILORS REFERENCING NEW TABLE NewSailors FOR EACH STATEMENT INSERT INTO YoungSailors(sid, name, age, rating) SELECT sid, name, age, rating FROM NewSailors N WHERE N.age <= 18 atabase Management Systems 3ed, R. Ramakrishnan and J. Gehrke 4

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