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SQL: DML and Constraints

SQL: DML and Constraints. Chapter 5. Review (1/3). In Chapter 3, we covered commands to create and modify tables: This included DDL commands: CREATE TABLE table_name (column1 domain 1, column2 domain 2, ... with some constraints) creates a new table

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SQL: DML and Constraints

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  1. SQL: DML and Constraints Chapter 5

  2. Review (1/3) • In Chapter 3, we covered commands to create and modify tables: • This included DDL commands: • CREATE TABLE table_name (column1 domain 1, column2 domain 2, ... with some constraints) • creates a new table • ALTER TABLE table_name ADD column ; • or ALTER TABLE table_name DROP column • Modifies tables • DROP TABLEtable_name • deletes a table

  3. Review (2/3) • In Chapter 3, we also covered commands to modify data in tables: • and Basic DML commands to modify tables: • UPDATE table SET attribute_values WHERE constraint: - updates existing data in a database • DELETE FROM table WHERE constraint: deletes tuples from a database • INSERT INTO table_name (column1, column2, column3,...) VALUES (value1, value2, value3,...)- inserts new data into a database • And simple SELECT commands

  4. Review 3/3 • In Chapter 3, we also covered basic constraints within CREATE TABLE: • PRIMARY KEY • UNIQUE • FOREIGN KEY • NOT NULL • CHECK • DEFAULT

  5. Introduction • In chapter 5, we cover advanced forms of SQL DML: • UNION, INTERSECT, EXCEPT • Nested Queries • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  6. Outline • Form of basic SQL: • Also with use of expressions and pattern matching • UNION, INTERSECT, EXCEPT • Nested Queries • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  7. R1 Example Instances S1 • We will use these instances of the Sailors and Reserves relations in our examples. S2

  8. BRIEF Basic SQL Query SELECT [DISTINCT] target-list FROMrelation-list WHERE 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!

  9. BRIEF 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.

  10. BRIEF Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103

  11. A Note on Range Variables • It is good style to use range variables (aliases) • Must be there if the same relation appears twice in the FROM clause. • Note that The previous query can also be written as: SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid ANDbid=103 OR SELECTsname FROM Sailors, Reserves WHERESailors.sid=Reserves.sid AND bid=103

  12. Example of basic SQL query:Find sailors who’ve reserved at least one boat SELECT S.sid FROM Sailors S, Reserves R WHERE S.sid=R.sid • Would adding DISTINCT to this query make a difference? • Note: Could be multiple reservations • How about adding DISTINCT matter for: SELECT S.sid FROM Sailors S • 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?

  13. A more general form of SQL SELECT [DISTINCT] target-list FROMrelation-list WHERE qualification • SQL supports having a target-list of the form: • Expression AS column_name; • Expression1 = Expression2 • Expression can can be any arithmetic or string expression over column names • Column_name becomes the new name for the resulting column

  14. Example1 : Expressions SELECT S.age, age1=S.age-5, 2*S.age ASage2 FROM Sailors S WHERE S.sname = ‘Hazem’ • Find triples (of ages of sailors and two fields defined by expressions) for sailors whose names = Hazem • Strings always included between single quotes • Illustrates use of arithmetic expressions • ASand = are two ways to name fields in result.

  15. Pattern Matching SELECT S.age, age1=S.age-5, 2*S.age ASage2 FROM Sailors S WHERE S.sname LIKE ‘G_%LE’ • SQL supports use of pattern matching. • LIKE is used for string matching. `_’ stands for any one character and `%’ stands for 0 or more arbitrary characters. • In above example, ‘G_%LE’ could be: GOOGLE, GIGGLE, GOGGLE, GOLE, GOALE (ignore spelling mistake ) ,…. • Example illustrates use string pattern matching: Find triples (of ages of sailors and two fields defined by expressions) for sailors whose names begin G and end with LE and contain at least four characters.

  16. Outline • Form of basic SQL • UNION, INTERSECT, EXCEPT • Nested Queries • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  17. Extensions to basic query • UNION: Can be used to compute the union of any two union-compatible sets of tuples (which are themselves the result of SQL queries). • INTERSECT: Can be used to compute the intersection of any two union-compatible sets of tuples. • EXCEPT: Can be used to compute the difference between two union-compatible sets of tuples. • Resulting set: No redundancy in tuples!

  18. Example 1: Find sid’s of sailors who’ve reserved a red or a green boat 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’) • If we replace ORby ANDin the first version, what do we get? • Note: No redundancy in tuples! • What do we get if we replace UNIONby 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’ UNION SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

  19. Example 2: Find sid’s of sailors who’ve reserved a red and a green boat SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color=‘red’ AND B2.color=‘green’) • What if I were looking for names instead of sids? • Would need to display both or use nested queries…see next SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ INTERSECT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

  20. Outline • Form of basic SQL • UNION, INTERSECT, EXCEPT • Nested Queries (IN, EXISTS, UNIQUE, op ANY, op ALL, NOT …) • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  21. Additional Extensions to basic query • Set operations used in Nested Queries: • IN: to check if an element in a given set • Op ANY: to compare a value to any element in the set • Op ALL: to compare a value to ALL elements in the set • EXISTS: to check if a set is empty • UNIQUE: to check if no row appears twice • NOT EXISTS, NOT IN, NOT UNIQUE

  22. Nested Queries – Use of IN Find 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) • A very powerful feature of SQL: a WHERE clause can itself contain an SQL query! • To find sailors who’ve 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.

  23. Nested Queries with Correlation - EXISTS Find names of sailors who’ve reserved boat #103: SELECT S.sname FROM Sailors S WHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid=103 ANDS.sid=R.sid) • EXISTSis another set comparison operator, like IN. • The use of * illustrates when we do not care what we find as long as we find something. • If UNIQUEis used instead of EXISTS, and * is replaced by R.bid, what do we get? • finds sailors with one reservation for boat #103. (UNIQUE checks for duplicate tuples; * denotes all attributes. We replace * by R.bid to focus on boat count)

  24. Find names of sailors who’ve reserved boat #103Sometime more than one way to say the same thing • Use basic construct: • SELECT S.sname • FROM Sailors S, Reserves R • WHERE S.sid=R.sid AND R.bid=103 • Use Nested Queries: • SELECT S.sname • FROM Sailors S • WHERE S.sid IN (SELECT R.sid • FROM Reserves R • WHERE R.bid=103) • Use Nested Queries with correlations: • SELECT S.sname • FROM Sailors S • WHERE EXISTS (SELECT * • FROM Reserves R • WHERE R.bid=103 ANDS.sid=R.sid)

  25. Example Set-Comparison Operators • 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’)

  26. Rewriting INTERSECT Queries Using IN Find sid’s of sailors who’ve reserved both a red and a green boat: SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE 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’) • Similarly, EXCEPT queries re-written using NOT IN. • To find names (not sid’s) of Sailors who’ve reserved both red and green boats, • just replace S.sid by S.sname in SELECT clause

  27. Division in SQL • Find sailors who’ve reserved all boats. • Hint: Find names of sailors for which • you can not find (not exist) • [boats that are not part of (except) reservations for this sailor] SELECT S.sname FROM Sailors S WHERE NOT EXISTS ((SELECT B.bid FROM Boats B) EXCEPT (SELECT R.bid FROM Reserves R WHERER.sid=S.sid)) (1)

  28. Division in SQL Find sailors who’ve reserved all boats. • We can also do it the hard way, without EXCEPT: Hint: Find Sailors S such that ... there is no boat B without ... a Reserves tuple showing S reserved B (2) SELECT S.sname FROM Sailors S WHERE NOT EXISTS (SELECT B.bid FROM Boats B WHERE NOT EXISTS (SELECT R.bid FROM Reserves R WHERER.bid=B.bid AND R.sid=S.sid))

  29. Outline • Form of basic SQL • UNION, INTERSECT, EXCEPT • Nested Queries • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  30. Aggregate Operators • An extension of relational algebra. COUNT (*) COUNT ( [DISTINCT] A) SUM ( [DISTINCT] A) AVG ( [DISTINCT] A) MAX (A) MIN (A) A: single column Most have the syntax: SELECT function (column_name ) FROM Sailors S

  31. Aggregate Operators • Find the average age of all sailors • SELECT AVG(S.age) FROM SAILORS S • Find the average age of sailors with a rating of 10. • SELECT AVG (S.age) FROM SAILORS S WHERE S.rating = 10 • Find the name and age of the oldest sailor • SELECT S. name , S. age FROM SAILORS S WHERE S.age = (SELECT MAX (S2.age) FROM SAILORS S2) • Count the number of sailors • SELECT COUNT (*) FROM SAILORS S • Count the number of different sailor names • SELECT COUNT ( DISNTINCT S.name) FROM SAILORS S • Find the names of sailors who are older than the oldest sailor with a rating of 10 SELECT S.name FROM SAILORS S WHERE S.age > (SELECT MAX (S2.age) FROM SAILORS S2 WHERE S2.rating = 10) SELECT S.sname, MAX (S.age) FROM Sailors S

  32. Stop here

  33. GROUP BY and HAVING • So far, we’ve 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 don’t 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 (!): SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2, ... , 10:

  34. Example: Find the age of the youngest sailor for each rating level SELECT S.rating, MIN (S.age) FROM Sailors S GROUP BY S.rating

  35. Queries With GROUP BY and HAVING SELECT [DISTINCT] target-list FROMrelation-list WHERE qualification GROUP BYgrouping-list HAVING group-qualification • Thetarget-listcontains • (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, andthese attributes must have a single value per group (ie common to all) • A group is a set of tuples that have the same value for all attributes in grouping-list.

  36. Conceptual Evaluation (in steps) • STEP 1: Cross-product relations in FROM… • STEP 2: Eliminate tuples that fail WHERE condition • STEP 3: Project into columns identified in SELECT, GROUP BY, HAVING • STEP 4: Sort the table according to GROUP BY • STEP 5: Eliminate groups that fail the condition in HAVING • STEP 6: Generate one answer row for remaining groups as specified in SELECT • STEP 7: Eliminate duplicates in the output if DISTINCT is specified

  37. Conceptual Evaluation (in English ) • 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-qualificationthat 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.

  38. Find the age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVINGCOUNT (*) > 1 • Only S.rating and S.age are mentioned in the SELECT, GROUP BY or HAVING clauses; other attributes `unnecessary’. • 2nd column of result is unnamed. (Use AS to name it.) Answer relation

  39. Prelim exercise • Find ratings that have more than two sailors. SELECTS.rating FROM Sailors S WHERE (1 < (SELECT COUNT (*) FROM Sailors S2 WHERES.rating=S2.rating))

  40. Find the age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age) • HAVING clause can also contain a subquery. SELECTS.rating, MIN (S.age) FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERES.rating=S2.rating)

  41. Find the age of the youngest sailor with age >= 18, for each rating with at least 2 such sailors(REVISITED) SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 SELECT Temp.rating, Temp.minage FROM ( Select S.rating, MIN (S.age) AS minage, COUNT (*) AS ratingcount FROM Sailors S WHERE S.age > 18 GROUP BY S.rating ) AS Temp WHERE Temp.ratingcount > 1

  42. For each red boat, find the number of reservations for this boat SELECT B.bid, COUNT (*) AS scount FROM Boats B, Reserves R WHERE R.bid=B.bid AND B.color=‘red’ GROUP BY B.bid

  43. Find the average age of sailors who are eligible to vote (over 18) for each rating level that has at least two sailors. • SELECT S.rating AVG(S.age) FROM Sailors S WHERE S.age >=18 GROUP BY S.rating HAVING (SELECT COUNT (*) FROM Sailors S2 S2.rating = S.rating) > 1

  44. Find the rating(s) for which the average age is the minimum over all ratings • In pseudo-SQL (English), we would want to say: SELECT S.rating FROM Sailors S WHERE AVG(S.age) = (SELECT MIN (AVG(S2.age)) FROM Sailors S2) But this is not supported by SQL (aggregations can not be nested.

  45. Find the rating(s) for which the average age is the minimum over all ratings • 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)

  46. Outline • Form of basic SQL • UNION, INTERSECT, EXCEPT • Nested Queries • Aggregate Operators • Null Values • Complex Integrity constraints • Triggers and Active Databases

  47. Null Values • Field values in a tuple are sometimes unknown(e.g., a rating has not been assigned) or inapplicable(e.g., no spouse’s 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). • Aggregations ignore null values, except COUNT(*). • Meaning of constructs must be defined carefully. (e.g., WHERE clause eliminates rows that don’t evaluate to true.) • outer joins uses null values in the join condition when the join condition is not satisfied

  48. 3-valued logic Basic SQL comparison operators always return Unknown when comparing anything with Null

  49. Outer Joins • Previously, Only considered (inner) join, where we select rows common to the participating tables to a join. • What about the cases where we are interested in selecting elements in a table regardless of whether they are present in the second table? • Use Outer Join • The syntax for performing an outer join in SQL is database-dependent.

  50. Typical Syntax • SELECT column_name(s)FROM table_name1LEFT OUTER JOIN table_name2ON table_name1.column_name=table_name2.column_name

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