SQL Data Manipulation

Week 2
SQL: Data Manipulation
Pearson Education © 2009

2
Learning Objectives
 Purpose and importance of SQL.
 How to retrieve data from database using
SELECT and:
– Use compound WHERE conditions.
– Sort query results using ORDER BY.
– Use aggregate functions.
– Group data using GROUP BY and HAVING.
– Use subqueries.

3
Learning Objectives
– Join tables together.
– Perform set operations (UNION, INTERSECT,
EXCEPT).
 How to update database using INSERT,
UPDATE, and DELETE.

4
Please Refer Text Book
Chapter 6
(Page 183-221)

5
Writing SQL Commands
 SQL statement consists of reserved words and user-
defined words.
– Reserved words are a fixed part of SQL and must
be spelt exactly as required and cannot be split
across lines.
– User-defined words are made up by user and
represent names of various database objects such
as relations, columns, views.

6
 Most components of an SQL statement are case
insensitive, except for literal character data.
 More readable with indentation and lineation:
– Each clause should begin on a new line.
– Start of a clause should line up with start of
other clauses.
– If clause has several parts, should each appear
on a separate line and be indented under start
of clause.

7
 Use extended form of BNF notation:
- Upper-case letters represent reserved words.
- Lower-case letters represent user-defined words.
- | indicates a choice among alternatives.
- Curly braces indicate a required element.
- Square brackets indicate an optional element.
- … indicates optional repetition (0 or more).

8
Literals
 Literals are constants used in SQL statements.
 All non-numeric literals must be enclosed in
single quotes (e.g. ‘London’).
 All numeric literals must not be enclosed in
quotes (e.g. 650.00).

9
SELECT Statement
SELECT [DISTINCT | ALL]
{* | [columnExpression [AS newName]] [,...] }
FROM TableName [alias] [, ...]
[WHERE condition]
[GROUP BY columnList] [HAVING condition]
[ORDER BY columnList]

10
SELECT Statement
FROM Specifies table(s) to be used.
WHERE Filters rows.
GROUP BY Forms groups of rows with same
column value.
HAVING Filters groups subject to some
condition.
SELECT Specifies which columns are to
appear in output.
ORDER BY Specifies the order of the output.

11
SELECT Statement
 Order of the clauses cannot be changed.
 Only SELECT and FROM are mandatory.

12
Example 6.1 All Columns, All Rows
List full details of all staff.
SELECT staffNo, fName, lName, address,
position, sex, DOB, salary, branchNo
FROM Staff;
 Can use * as an abbreviation for ‘all columns’:
SELECT *
FROM Staff;

13
Example 6.1 All Columns, All Rows

14
Example 6.2 Specific Columns, All Rows
Produce a list of salaries for all staff, showing only
staff number, first and last names, and salary.
SELECT staffNo, fName, lName, salary
FROM Staff;

15
Example 6.2 Specific Columns, All Rows

16
Example 6.3 Use of DISTINCT
List the property numbers of all properties that
have been viewed.
SELECT propertyNo
FROM Viewing;

17
Example 6.3 Use of DISTINCT
 Use DISTINCT to eliminate duplicates:
SELECT DISTINCT propertyNo
FROM Viewing;

18
Example 6.4 Calculated Fields
Produce list of monthly salaries for all staff,
showing staff number, first/last name, and salary.
SELECT staffNo, fName, lName, salary/12
FROM Staff;

19
Example 6.4 Calculated Fields
 To name column, use AS clause:
SELECT staffNo, fName, lName, salary/12
AS monthlySalary
FROM Staff;

20
Example 6.5 Comparison Search Condition
List all staff with a salary greater than 10,000.
SELECT staffNo, fName, lName, position, salary
FROM Staff
WHERE salary > 10000;

21
Example 6.6 Compound Comparison Search
Condition
List addresses of all branch offices in London or
Glasgow.
SELECT *
FROM Branch
WHERE city = ‘London’ OR city = ‘Glasgow’;

22
Example 6.7 Range Search Condition
List all staff with a salary between 20,000 and
30,000.
FROM Staff
WHERE salary BETWEEN 20000 AND 30000;
 BETWEEN test includes the endpoints of range.

23

24
 Also a negated version NOT BETWEEN.
 BETWEEN does not add much to SQL’s
expressive power. Could also write:
FROM Staff
WHERE salary>=20000 AND salary <= 30000;
 Useful, though, for a range of values.

25
Example 6.8 Set Membership
List all managers and supervisors.
SELECT staffNo, fName, lName, position
FROM Staff
WHERE position IN (‘Manager’, ‘Supervisor’);

26
Example 6.8 Set Membership
 There is a negated version (NOT IN).
 IN does not add much to SQL’s expressive power.
Could have expressed this as:
FROM Staff
WHERE position=‘Manager’ OR
position=‘Supervisor’;
 IN is more efficient when set contains many values.

27
Example 6.9 Pattern Matching
Find all owners with the string ‘Glasgow’ in their
address.
SELECT ownerNo, fName, lName, address, telNo
FROM PrivateOwner
WHERE address LIKE ‘%Glasgow%’;

28
Example 6.9 Pattern Matching
 SQL has two special pattern matching symbols:
– %: sequence of zero or more characters;
– _ (underscore): any single character.
 LIKE ‘%Glasgow%’ means a sequence of
characters of any length containing ‘Glasgow’.

29
Example 6.10 NULL Search Condition
List details of all viewings on property PG4
where a comment has not been supplied.
 There are 2 viewings for property PG4, one with
and one without a comment.
 Have to test for null explicitly using special
keyword IS NULL:
SELECT clientNo, viewDate
FROM Viewing
WHERE propertyNo = ‘PG4’ AND
comment IS NULL;

30
Example 6.10 NULL Search Condition
 Negated version (IS NOT NULL) can test for
non-null values.

31
Example 6.11 Single Column Ordering
List salaries for all staff, arranged in descending
order of salary.
SELECT staffNo, fName, lName, salary
FROM Staff
ORDER BY salary DESC;

32
Example 6.11 Single Column Ordering

33
Example 6.12 Multiple Column Ordering
Produce abbreviated list of properties in order of
property type.
SELECT propertyNo, type, rooms, rent
FROM PropertyForRent
ORDER BY type;

34

35
 Four flats in this list - as no minor sort key
specified, system arranges these rows in any order
it chooses.
 To arrange in order of rent, specify minor order:
SELECT propertyNo, type, rooms, rent
ORDER BY type, rent DESC;

36

37
SELECT Statement - Aggregates
 ISO standard defines five aggregate functions:
COUNT returns number of values in specified
column.
SUM returns sum of values in specified column.
AVG returns average of values in specified column.
MIN returns smallest value in specified column.
MAX returns largest value in specified column.

38
 Each operates on a single column of a table and
returns a single value.
 COUNT, MIN, and MAX apply to numeric and
non-numeric fields, but SUM and AVG may be
used on numeric fields only.
 Apart from COUNT(*), each function eliminates
nulls first and operates only on remaining non-
null values.

39
 COUNT(*) counts all rows of a table, regardless
of whether nulls or duplicate values occur.
 Can use DISTINCT before column name to
eliminate duplicates.
 DISTINCT has no effect with MIN/MAX, but
may have with SUM/AVG.

40
 Aggregate functions can be used only in
SELECT list and in HAVING clause.
 If SELECT list includes an aggregate function
and there is no GROUP BY clause, SELECT list
cannot reference a column out with an aggregate
function. For example, the following is illegal:
SELECT staffNo, COUNT(salary)
FROM Staff;

41
Example 6.13 Use of COUNT(*)
How many properties cost more than £350 per
month to rent?
SELECT COUNT(*) AS myCount
WHERE rent > 350;

42
Example 6.14 Use of COUNT(DISTINCT)
How many different properties viewed in May ‘04?
SELECT COUNT(DISTINCT propertyNo) AS myCount
FROM Viewing
WHERE viewDate BETWEEN ‘1-May-04’
AND ‘31-May-04’;

43
Example 6.15 Use of COUNT and SUM
Find number of Managers and sum of their
salaries.
SELECT COUNT(staffNo) AS myCount,
SUM(salary) AS mySum
FROM Staff
WHERE position = ‘Manager’;

44
Example 6.16 Use of MIN, MAX, AVG
Find minimum, maximum, and average staff
salary.
SELECT MIN(salary) AS myMin,
MAX(salary) AS myMax,
AVG(salary) AS myAvg
FROM Staff;

45
SELECT Statement - Grouping
 Use GROUP BY clause to get sub-totals.
 SELECT and GROUP BY closely integrated:
each item in SELECT list must be single-valued
per group, and SELECT clause may only contain:
– column names
– aggregate functions
– constants
– expression involving combinations of the above.

46
SELECT Statement - Grouping
 All column names in SELECT list must appear in
GROUP BY clause unless name is used only in an
aggregate function.
 If WHERE is used with GROUP BY, WHERE is
applied first, then groups are formed from
remaining rows satisfying predicate.
 ISO considers two nulls to be equal for purposes
of GROUP BY.

47
Example 6.17 Use of GROUP BY
Find number of staff in each branch and their
total salaries.
SELECT branchNo,
COUNT(staffNo) AS myCount,
FROM Staff
GROUP BY branchNo
ORDER BY branchNo;

48
Example 6.17 Use of GROUP BY

49
Restricted Groupings – HAVING clause
 HAVING clause is designed for use with GROUP
BY to restrict groups that appear in final result
table.
 Similar to WHERE, but WHERE filters
individual rows whereas HAVING filters groups.
 Column names in HAVING clause must also
appear in the GROUP BY list or be contained
within an aggregate function.

50
Example 6.18 Use of HAVING
For each branch with more than 1 member of
staff, find number of staff in each branch and
sum of their salaries.
SELECT branchNo,
COUNT(staffNo) AS myCount,
FROM Staff
GROUP BY branchNo
HAVING COUNT(staffNo) > 1
ORDER BY branchNo;

51
Example 6.18 Use of HAVING

52
Subqueries
 Some SQL statements can have a SELECT
embedded within them.
 A subselect can be used in WHERE and
HAVING clauses of an outer SELECT, where it
is called a subquery or nested query.
 Subselects may also appear in INSERT,
UPDATE, and DELETE statements.

53
Example 6.19 Subquery with Equality
List staff who work in branch at ‘163 Main St’.
FROM Staff
WHERE branchNo =
(SELECT branchNo
FROM Branch
WHERE street = ‘163 Main St’);

54
 Inner SELECT finds branch number for branch
at ‘163 Main St’ (‘B003’).
 Outer SELECT then retrieves details of all staff
who work at this branch.
 Outer SELECT then becomes:
FROM Staff
WHERE branchNo = ‘B003’;

55

56
Example 6.20 Subquery with Aggregate
List all staff whose salary is greater than the average
salary, and show by how much.
SELECT staffNo, fName, lName, position,
salary – (SELECT AVG(salary) FROM Staff) As SalDiff
FROM Staff
WHERE salary >
(SELECT AVG(salary)
FROM Staff);

57
 Cannot write ‘WHERE salary > AVG(salary)’
 Instead, use subquery to find average salary
(17000), and then use outer SELECT to find those
staff with salary greater than this:
SELECT staffNo, fName, lName, position,
salary – 17000 As salDiff
FROM Staff
WHERE salary > 17000;

58

59
Subquery Rules
 ORDER BY clause may not be used in a
subquery (although it may be used in outermost
SELECT).
 Subquery SELECT list must consist of a single
column name or expression, except for
subqueries that use EXISTS.
 By default, column names refer to table name in
FROM clause of subquery. Can refer to a table
in FROM using an alias.

60
Subquery Rules
 When subquery is an operand in a comparison,
subquery must appear on right-hand side.
 A subquery may not be used as an operand in an
expression.

61
Example 6.21 Nested subquery: use of IN
List properties handled by staff at ‘163 Main St’.
SELECT propertyNo, street, city, postcode, type, rooms, rent
WHERE staffNo IN
(SELECT staffNo
FROM Staff
WHERE branchNo =
(SELECT branchNo
FROM Branch
WHERE street = ‘163 Main St’));

62
Example 6.21 Nested subquery: use of IN

63
ANY and ALL
 ANY and ALL may be used with subqueries that
produce a single column of numbers.
 With ALL, condition will only be true if it is
satisfied by all values produced by subquery.
 With ANY, condition will be true if it is satisfied
by any values produced by subquery.
 If subquery is empty, ALL returns true, ANY
returns false.
 SOME may be used in place of ANY.

64
Example 6.22 Use of ANY/SOME
Find staff whose salary is larger than salary of at
least one member of staff at branch B003.
FROM Staff
WHERE salary > SOME
(SELECT salary
FROM Staff
WHERE branchNo = ‘B003’);

65
Example 6.22 Use of ANY/SOME
 Inner query produces set {12000, 18000, 24000}
and outer query selects those staff whose salaries
are greater than any of the values in this set.

66
Example 6.23 Use of ALL
Find staff whose salary is larger than salary of
every member of staff at branch B003.
FROM Staff
WHERE salary > ALL
(SELECT salary
FROM Staff
WHERE branchNo = ‘B003’);

67
Example 6.23 Use of ALL

68
Multi-Table Queries
 Can use subqueries provided result columns come
from same table.
 If result columns come from more than one table
must use a join.
 To perform join, include more than one table in
FROM clause.
 Use comma as separator and typically include
WHERE clause to specify join column(s).

69
Multi-Table Queries
 Also possible to use an alias for a table named in
FROM clause.
 Alias is separated from table name with a space.
 Alias can be used to qualify column names when
there is ambiguity.

70
Example 6.24 Simple Join
List names of all clients who have viewed a
property along with any comment supplied.
SELECT c.clientNo, fName, lName,
propertyNo, comment
FROM Client c, Viewing v
WHERE c.clientNo = v.clientNo;

71
Example 6.24 Simple Join
 Only those rows from both tables that have
identical values in the clientNo columns
(c.clientNo = v.clientNo) are included in result.
 Equivalent to equi-join in relational algebra.

72
Alternative JOIN Constructs
 SQL provides alternative ways to specify joins:
FROM Client c JOIN Viewing v ON c.clientNo = v.clientNo
FROM Client JOIN Viewing USING clientNo
FROM Client NATURAL JOIN Viewing
 In each case, FROM replaces original FROM and
WHERE. However, first produces table with two
identical clientNo columns.

73
Example 6.25 Sorting a join
For each branch, list numbers and names of
staff who manage properties, and properties
they manage.
SELECT s.branchNo, s.staffNo, fName, lName,
propertyNo
FROM Staff s, PropertyForRent p
WHERE s.staffNo = p.staffNo
ORDER BY s.branchNo, s.staffNo, propertyNo;

74
Example 6.25 Sorting a join

75
Example 6.26 Three Table Join
For each branch, list staff who manage
properties, including city in which branch is
located and properties they manage.
SELECT b.branchNo, b.city, s.staffNo, fName, lName,
propertyNo
FROM Branch b, Staff s, PropertyForRent p
WHERE b.branchNo = s.branchNo AND
s.staffNo = p.staffNo
ORDER BY b.branchNo, s.staffNo, propertyNo;

76
Example 6.26 Three Table Join
 Alternative formulation for FROM and WHERE:
FROM (Branch b JOIN Staff s USING branchNo) AS
bs JOIN PropertyForRent p USING staffNo

77
Example 6.27 Multiple Grouping Columns
Find number of properties handled by each staff
member.
SELECT s.branchNo, s.staffNo, COUNT(*) AS myCount
GROUP BY s.branchNo, s.staffNo
ORDER BY s.branchNo, s.staffNo;

78
Example 6.27 Multiple Grouping Columns

79
Computing a Join
Procedure for generating results of a join are:
1. Form Cartesian product of the tables named in
FROM clause.
2. If there is a WHERE clause, apply the search
condition to each row of the product table,
retaining those rows that satisfy the condition.
3. For each remaining row, determine value of each
item in SELECT list to produce a single row in
result table.

80
Computing a Join
4. If DISTINCT has been specified, eliminate any
duplicate rows from the result table.
6. If there is an ORDER BY clause, sort result table
as required.
 SQL provides special format of SELECT for
Cartesian product:
SELECT [DISTINCT | ALL] {* | columnList}
FROM Table1 CROSS JOIN Table2

81
Outer Joins
If one row of a joined table is unmatched,
row is omitted from result table.
Outer join operations retain rows that do
not satisfy the join condition.
Consider following tables:

82
Outer Joins
 The (inner) join of these two tables:
SELECT b.*, p.*
FROM Branch1 b, PropertyForRent1 p
WHERE b.bCity = p.pCity;

83
Outer Joins
 Result table has two rows where cities are same.
 There are no rows corresponding to branches in
Bristol and Aberdeen.
 To include unmatched rows in result table, use
an Outer join.

84
Example 6.28 Left Outer Join
List branches and properties that are in same
city along with any unmatched branches.
SELECT b.*, p.*
FROM Branch1 b LEFT JOIN
PropertyForRent1 p ON b.bCity = p.pCity;

85
Example 6.28 Left Outer Join
 Includes those rows of first (left) table unmatched
with rows from second (right) table.
 Columns from second table are filled with
NULLs.

86
Example 6.29 Right Outer Join
List branches and properties in same city and
any unmatched properties.
SELECT b.*, p.*
FROM Branch1 b RIGHT JOIN

87
Example 6.29 Right Outer Join
 Right Outer join includes those rows of second
(right) table that are unmatched with rows from
first (left) table.
 Columns from first table are filled with NULLs.

88
Example 6.30 Full Outer Join
List branches and properties in same city and
any unmatched branches or properties.
SELECT b.*, p.*
FROM Branch1 b FULL JOIN

89
Example 6.30 Full Outer Join
 Includes rows that are unmatched in both tables.
 Unmatched columns are filled with NULLs.

90
EXISTS and NOT EXISTS
 EXISTS and NOT EXISTS are for use only with
subqueries.
 Produce a simple true/false result.
 True if and only if there exists at least one row in
result table returned by subquery.
 False if subquery returns an empty result table.
 NOT EXISTS is the opposite of EXISTS.

91
EXISTS and NOT EXISTS
 As (NOT) EXISTS check only for existence or non-
existence of rows in subquery result table,
subquery can contain any number of columns.
 Common for subqueries following (NOT) EXISTS
to be of form:
(SELECT * ...)

92
Example 6.31 Query using EXISTS
Find all staff who work in a London branch.
FROM Staff s
WHERE EXISTS
(SELECT *
FROM Branch b
WHERE s.branchNo = b.branchNo AND
city = ‘London’);

93

94
 Note, search condition s.branchNo = b.branchNo
is necessary to consider correct branch record for
each member of staff.
 If omitted, would get all staff records listed out
because subquery:
SELECT * FROM Branch WHERE city=‘London’
 would always be true and query would be:
SELECT staffNo, fName, lName, position FROM Staff
WHERE true;

95
 Could also write this query using join construct:
FROM Staff s, Branch b
WHERE s.branchNo = b.branchNo AND
city = ‘London’;

96
Union, Intersect, and Difference (Except)
 Can use normal set operations of Union,
Intersection, and Difference to combine results of
two or more queries into a single result table.
 Union of two tables, A and B, is table containing
all rows in either A or B or both.
 Intersection is table containing all rows common
to both A and B.
 Difference is table containing all rows in A but
not in B.
 Two tables must be union compatible.

97
 Format of set operator clause in each case is:
op [ALL] [CORRESPONDING [BY {column1 [, ...]}]]
 If CORRESPONDING BY specified, set operation
performed on the named column(s).
 If CORRESPONDING specified but not BY
clause, operation performed on common columns.
 If ALL specified, result can include duplicate rows.

98

99
Example 6.32 Use of UNION
List all cities where there is either a branch office
or a property.
(SELECT city
FROM Branch
WHERE city IS NOT NULL) UNION
(SELECT city
WHERE city IS NOT NULL);

100
 Or
(SELECT *
FROM Branch
WHERE city IS NOT NULL)
UNION CORRESPONDING BY city
(SELECT *
WHERE city IS NOT NULL);

101
 Produces result tables from both queries and
merges both tables together.

102
Example 6.33 Use of INTERSECT
List all cities where there is both a branch office
and a property.
(SELECT city FROM Branch)
INTERSECT
(SELECT city FROM PropertyForRent);

103
 Or
(SELECT * FROM Branch)
INTERSECT CORRESPONDING BY city
(SELECT * FROM PropertyForRent);

104
 Could rewrite this query without INTERSECT
operator:
SELECT b.city
FROM Branch b PropertyForRent p
WHERE b.city = p.city;
 Or:
SELECT DISTINCT city FROM Branch b
WHERE EXISTS
(SELECT * FROM PropertyForRent p
WHERE p.city = b.city);

105
Example 6.34 Use of EXCEPT
List of all cities where there is a branch office but
no properties.
(SELECT city FROM Branch)
EXCEPT
 Or
(SELECT * FROM Branch)
EXCEPT CORRESPONDING BY city
(SELECT * FROM PropertyForRent);

106
Example 6.34 Use of EXCEPT
 Could rewrite this query without EXCEPT:
SELECT DISTINCT city FROM Branch
WHERE city NOT IN
 Or
SELECT DISTINCT city FROM Branch b
WHERE NOT EXISTS
(SELECT * FROM PropertyForRent p
WHERE p.city = b.city);

107
INSERT
INSERT INTO TableName [ (columnList) ]
VALUES (dataValueList)
 columnList is optional; if omitted, SQL assumes a
list of all columns in their original CREATE
TABLE order.
 Any columns omitted must have been declared as
NULL when table was created, unless DEFAULT
was specified when creating column.

108
INSERT
 dataValueList must match columnList as follows:
– number of items in each list must be same;
– must be direct correspondence in position of
items in two lists;
– data type of each item in dataValueList must
be compatible with data type of
corresponding column.

109
Example 6.35 INSERT … VALUES
Insert a new row into Staff table supplying data
for all columns.
INSERT INTO Staff
VALUES (‘SG16’, ‘Alan’, ‘Brown’, ‘Assistant’,
‘M’, Date‘1957-05-25’, 8300, ‘B003’);

110
Example 6.36 INSERT using Defaults
Insert a new row into Staff table supplying data
for all mandatory columns.
INSERT INTO Staff (staffNo, fName, lName,
position, salary, branchNo)
VALUES (‘SG44’, ‘Anne’, ‘Jones’,
‘Assistant’, 8100, ‘B003’);
 Or
INSERT INTO Staff
VALUES (‘SG44’, ‘Anne’, ‘Jones’, ‘Assistant’, NULL,
NULL, 8100, ‘B003’);

111
INSERT … SELECT
 Second form of INSERT allows multiple rows to
be copied from one or more tables to another:
INSERT INTO TableName [ (columnList) ]
SELECT ...

112
Example 6.37 INSERT … SELECT
Assume there is a table StaffPropCount that
contains names of staff and number of properties
they manage:
StaffPropCount(staffNo, fName, lName, propCnt)
Populate StaffPropCount using Staff and
PropertyForRent tables.

113
INSERT INTO StaffPropCount
(SELECT s.staffNo, fName, lName, COUNT(*)
GROUP BY s.staffNo, fName, lName)
UNION
(SELECT staffNo, fName, lName, 0
FROM Staff
WHERE staffNo NOT IN
(SELECT DISTINCT staffNo
FROM PropertyForRent));

114
 If second part of UNION is omitted, excludes those
staff who currently do not manage any properties.

115
UPDATE
UPDATE TableName
SET columnName1 = dataValue1
[, columnName2 = dataValue2...]
[WHERE searchCondition]
 TableName can be name of a base table or an
updatable view.
 SET clause specifies names of one or more
columns that are to be updated.

116
UPDATE
 WHERE clause is optional:
– if omitted, named columns are updated for all
rows in table;
– if specified, only those rows that satisfy
searchCondition are updated.
 New dataValue(s) must be compatible with data
type for corresponding column.

117
Example 6.38/39 UPDATE All Rows
Give all staff a 3% pay increase.
UPDATE Staff
SET salary = salary*1.03;
Give all Managers a 5% pay increase.
UPDATE Staff
SET salary = salary*1.05
WHERE position = ‘Manager’;

118
Example 6.40 UPDATE Multiple Columns
Promote David Ford (staffNo=‘SG14’) to
Manager and change his salary to £18,000.
UPDATE Staff
SET position = ‘Manager’, salary = 18000
WHERE staffNo = ‘SG14’;

119
DELETE
DELETE FROM TableName
[WHERE searchCondition]
 TableName can be name of a base table or an
updatable view.
 searchCondition is optional; if omitted, all rows
are deleted from table. This does not delete table.
If search_condition is specified, only those rows
that satisfy condition are deleted.

120
Example 6.41/42 DELETE Specific Rows
Delete all viewings that relate to property PG4.
DELETE FROM Viewing
WHERE propertyNo = ‘PG4’;
Delete all records from the Viewing table.
DELETE FROM Viewing;

SQL Data Manipulation

Recommended

More Related Content

What's hot (20)

Viewers also liked (20)

Similar to SQL Data Manipulation (20)

Recently uploaded (20)

SQL Data Manipulation

Editor's Notes